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  • [Java]Queue in while loop, cannot modify the value?

    - by javaLearner.java
    This is my code: Iterator it = queue.iterator(); while(it.hasNext()){ random = randNumber(1,2); if(random == 1){ queue.poll(); } else { queue.add("new"); queue.poll(); } } It gives me: Exception in thread "test" java.util.ConcurrentModificationException at java.util.LinkedList$ListItr.checkForComodification(LinkedList.java:761) at java.util.LinkedList$ListItr.next(LinkedList.java:696) Edit @Jon Skeet: What I want to do is: I have a queue list in, let say the size is 10, lets say: a,b,c,d ... j Generate a number between 1 and 2. if 1, pull (remove the top element) else if 2 add new element I will stop the loop until I added 3 new elements

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  • Java Util Linked List - how to find next?

    - by drozzy
    When using Java LinkedList how do you find out the element's next or previous relationships? I mean, in a regular linked list I would do something like this: Node node1 = new Node(); Node node2 = new Node(); LinkedList list = new LinkedList(); list.add(node1); list.add(node2); //then my node1 will know who it's next is: assertEquals(node2, node1.next()); But in Java's LinkedList, the data does not seem to be modified. So how do I actually find out who the "next" (or "previous" in the case of doubly-linked lists) element is?

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  • Cannot create an array of LinkedLists in Java...?

    - by kchau
    I'm working on a sparse matrix class that needs to use an array of LinkedLists to store the values of a matrix. Each element of the array (i.e. each LinkedList) represents a row of the matrix. And, each element in the LinkedLists represents a column and the stored value. In my class, I have a declaration of the array as: private LinkedList<IntegerNode>[] myMatrix; And, in my constructor for the SparseMatrix, I try to define: myMatrix = new LinkedList<IntegerNode>[numRows]; The error I end up getting is "Cannot create a generic array of LinkedList<IntegerNode>." So, I have two issues with this, 1) What am I doing wrong, and 2) Why is the type acceptable in the declaration for the array if it can't be created? Edit: IntegerNode is a class that I have created. And, all of my class files are packaged together.

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  • convert object to in

    - by hoora
    hello! i'm beginner and i want to write a java code in eclipse! this program take two linkedlist of integers(for exp:a & b) and make a linkedlist (for exp:d) that every elements of it are summation of elements of that linkedlist! but i can't add this two element of linkedlist because these are Object!! please help me!! exp: a=[3,4,6,7,8] b=[4,3,7,5,3,2,1] d=[7,7,13,12,11,2,1] THANK YOU VERY VERY VERY MUCH!

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  • Optimizing AES modes on Solaris for Intel Westmere

    - by danx
    Optimizing AES modes on Solaris for Intel Westmere Review AES is a strong method of symmetric (secret-key) encryption. It is a U.S. FIPS-approved cryptographic algorithm (FIPS 197) that operates on 16-byte blocks. AES has been available since 2001 and is widely used. However, AES by itself has a weakness. AES encryption isn't usually used by itself because identical blocks of plaintext are always encrypted into identical blocks of ciphertext. This encryption can be easily attacked with "dictionaries" of common blocks of text and allows one to more-easily discern the content of the unknown cryptotext. This mode of encryption is called "Electronic Code Book" (ECB), because one in theory can keep a "code book" of all known cryptotext and plaintext results to cipher and decipher AES. In practice, a complete "code book" is not practical, even in electronic form, but large dictionaries of common plaintext blocks is still possible. Here's a diagram of encrypting input data using AES ECB mode: Block 1 Block 2 PlainTextInput PlainTextInput | | | | \/ \/ AESKey-->(AES Encryption) AESKey-->(AES Encryption) | | | | \/ \/ CipherTextOutput CipherTextOutput Block 1 Block 2 What's the solution to the same cleartext input producing the same ciphertext output? The solution is to further process the encrypted or decrypted text in such a way that the same text produces different output. This usually involves an Initialization Vector (IV) and XORing the decrypted or encrypted text. As an example, I'll illustrate CBC mode encryption: Block 1 Block 2 PlainTextInput PlainTextInput | | | | \/ \/ IV >----->(XOR) +------------->(XOR) +---> . . . . | | | | | | | | \/ | \/ | AESKey-->(AES Encryption) | AESKey-->(AES Encryption) | | | | | | | | | \/ | \/ | CipherTextOutput ------+ CipherTextOutput -------+ Block 1 Block 2 The steps for CBC encryption are: Start with a 16-byte Initialization Vector (IV), choosen randomly. XOR the IV with the first block of input plaintext Encrypt the result with AES using a user-provided key. The result is the first 16-bytes of output cryptotext. Use the cryptotext (instead of the IV) of the previous block to XOR with the next input block of plaintext Another mode besides CBC is Counter Mode (CTR). As with CBC mode, it also starts with a 16-byte IV. However, for subsequent blocks, the IV is just incremented by one. Also, the IV ix XORed with the AES encryption result (not the plain text input). Here's an illustration: Block 1 Block 2 PlainTextInput PlainTextInput | | | | \/ \/ AESKey-->(AES Encryption) AESKey-->(AES Encryption) | | | | \/ \/ IV >----->(XOR) IV + 1 >---->(XOR) IV + 2 ---> . . . . | | | | \/ \/ CipherTextOutput CipherTextOutput Block 1 Block 2 Optimization Which of these modes can be parallelized? ECB encryption/decryption can be parallelized because it does more than plain AES encryption and decryption, as mentioned above. CBC encryption can't be parallelized because it depends on the output of the previous block. However, CBC decryption can be parallelized because all the encrypted blocks are known at the beginning. CTR encryption and decryption can be parallelized because the input to each block is known--it's just the IV incremented by one for each subsequent block. So, in summary, for ECB, CBC, and CTR modes, encryption and decryption can be parallelized with the exception of CBC encryption. How do we parallelize encryption? By interleaving. Usually when reading and writing data there are pipeline "stalls" (idle processor cycles) that result from waiting for memory to be loaded or stored to or from CPU registers. Since the software is written to encrypt/decrypt the next data block where pipeline stalls usually occurs, we can avoid stalls and crypt with fewer cycles. This software processes 4 blocks at a time, which ensures virtually no waiting ("stalling") for reading or writing data in memory. Other Optimizations Besides interleaving, other optimizations performed are Loading the entire key schedule into the 128-bit %xmm registers. This is done once for per 4-block of data (since 4 blocks of data is processed, when present). The following is loaded: the entire "key schedule" (user input key preprocessed for encryption and decryption). This takes 11, 13, or 15 registers, for AES-128, AES-192, and AES-256, respectively The input data is loaded into another %xmm register The same register contains the output result after encrypting/decrypting Using SSSE 4 instructions (AESNI). Besides the aesenc, aesenclast, aesdec, aesdeclast, aeskeygenassist, and aesimc AESNI instructions, Intel has several other instructions that operate on the 128-bit %xmm registers. Some common instructions for encryption are: pxor exclusive or (very useful), movdqu load/store a %xmm register from/to memory, pshufb shuffle bytes for byte swapping, pclmulqdq carry-less multiply for GCM mode Combining AES encryption/decryption with CBC or CTR modes processing. Instead of loading input data twice (once for AES encryption/decryption, and again for modes (CTR or CBC, for example) processing, the input data is loaded once as both AES and modes operations occur at in the same function Performance Everyone likes pretty color charts, so here they are. I ran these on Solaris 11 running on a Piketon Platform system with a 4-core Intel Clarkdale processor @3.20GHz. Clarkdale which is part of the Westmere processor architecture family. The "before" case is Solaris 11, unmodified. Keep in mind that the "before" case already has been optimized with hand-coded Intel AESNI assembly. The "after" case has combined AES-NI and mode instructions, interleaved 4 blocks at-a-time. « For the first table, lower is better (milliseconds). The first table shows the performance improvement using the Solaris encrypt(1) and decrypt(1) CLI commands. I encrypted and decrypted a 1/2 GByte file on /tmp (swap tmpfs). Encryption improved by about 40% and decryption improved by about 80%. AES-128 is slighty faster than AES-256, as expected. The second table shows more detail timings for CBC, CTR, and ECB modes for the 3 AES key sizes and different data lengths. » The results shown are the percentage improvement as shown by an internal PKCS#11 microbenchmark. And keep in mind the previous baseline code already had optimized AESNI assembly! The keysize (AES-128, 192, or 256) makes little difference in relative percentage improvement (although, of course, AES-128 is faster than AES-256). Larger data sizes show better improvement than 128-byte data. Availability This software is in Solaris 11 FCS. It is available in the 64-bit libcrypto library and the "aes" Solaris kernel module. You must be running hardware that supports AESNI (for example, Intel Westmere and Sandy Bridge, microprocessor architectures). The easiest way to determine if AES-NI is available is with the isainfo(1) command. For example, $ isainfo -v 64-bit amd64 applications pclmulqdq aes sse4.2 sse4.1 ssse3 popcnt tscp ahf cx16 sse3 sse2 sse fxsr mmx cmov amd_sysc cx8 tsc fpu 32-bit i386 applications pclmulqdq aes sse4.2 sse4.1 ssse3 popcnt tscp ahf cx16 sse3 sse2 sse fxsr mmx cmov sep cx8 tsc fpu No special configuration or setup is needed to take advantage of this software. Solaris libraries and kernel automatically determine if it's running on AESNI-capable machines and execute the correctly-tuned software for the current microprocessor. Summary Maximum throughput of AES cipher modes can be achieved by combining AES encryption with modes processing, interleaving encryption of 4 blocks at a time, and using Intel's wide 128-bit %xmm registers and instructions. References "Block cipher modes of operation", Wikipedia Good overview of AES modes (ECB, CBC, CTR, etc.) "Advanced Encryption Standard", Wikipedia "Current Modes" describes NIST-approved block cipher modes (ECB,CBC, CFB, OFB, CCM, GCM)

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  • 8-Puzzle Solution executes infinitely [migrated]

    - by Ashwin
    I am looking for a solution to 8-puzzle problem using the A* Algorithm. I found this project on the internet. Please see the files - proj1 and EightPuzzle. The proj1 contains the entry point for the program(the main() function) and EightPuzzle describes a particular state of the puzzle. Each state is an object of the 8-puzzle. I feel that there is nothing wrong in the logic. But it loops forever for these two inputs that I have tried : {8,2,7,5,1,6,3,0,4} and {3,1,6,8,4,5,7,2,0}. Both of them are valid input states. What is wrong with the code? Note For better viewing copy the code in a Notepad++ or some other text editor(which has the capability to recognize java source file) because there are lot of comments in the code. Since A* requires a heuristic, they have provided the option of using manhattan distance and a heuristic that calculates the number of misplaced tiles. And to ensure that the best heuristic is executed first, they have implemented a PriorityQueue. The compareTo() function is implemented in the EightPuzzle class. The input to the program can be changed by changing the value of p1d in the main() function of proj1 class. The reason I am telling that there exists solution for the two my above inputs is because the applet here solves them. Please ensure that you select 8-puzzle from teh options in the applet. EDITI gave this input {0,5,7,6,8,1,2,4,3}. It took about 10 seconds and gave a result with 26 moves. But the applet gave a result with 24 moves in 0.0001 seconds with A*. For quick reference I have pasted the the two classes without the comments : EightPuzzle import java.util.*; public class EightPuzzle implements Comparable <Object> { int[] puzzle = new int[9]; int h_n= 0; int hueristic_type = 0; int g_n = 0; int f_n = 0; EightPuzzle parent = null; public EightPuzzle(int[] p, int h_type, int cost) { this.puzzle = p; this.hueristic_type = h_type; this.h_n = (h_type == 1) ? h1(p) : h2(p); this.g_n = cost; this.f_n = h_n + g_n; } public int getF_n() { return f_n; } public void setParent(EightPuzzle input) { this.parent = input; } public EightPuzzle getParent() { return this.parent; } public int inversions() { /* * Definition: For any other configuration besides the goal, * whenever a tile with a greater number on it precedes a * tile with a smaller number, the two tiles are said to be inverted */ int inversion = 0; for(int i = 0; i < this.puzzle.length; i++ ) { for(int j = 0; j < i; j++) { if(this.puzzle[i] != 0 && this.puzzle[j] != 0) { if(this.puzzle[i] < this.puzzle[j]) inversion++; } } } return inversion; } public int h1(int[] list) // h1 = the number of misplaced tiles { int gn = 0; for(int i = 0; i < list.length; i++) { if(list[i] != i && list[i] != 0) gn++; } return gn; } public LinkedList<EightPuzzle> getChildren() { LinkedList<EightPuzzle> children = new LinkedList<EightPuzzle>(); int loc = 0; int temparray[] = new int[this.puzzle.length]; EightPuzzle rightP, upP, downP, leftP; while(this.puzzle[loc] != 0) { loc++; } if(loc % 3 == 0){ temparray = this.puzzle.clone(); temparray[loc] = temparray[loc + 1]; temparray[loc + 1] = 0; rightP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); rightP.setParent(this); children.add(rightP); }else if(loc % 3 == 1){ //add one child swaps with right temparray = this.puzzle.clone(); temparray[loc] = temparray[loc + 1]; temparray[loc + 1] = 0; rightP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); rightP.setParent(this); children.add(rightP); //add one child swaps with left temparray = this.puzzle.clone(); temparray[loc] = temparray[loc - 1]; temparray[loc - 1] = 0; leftP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); leftP.setParent(this); children.add(leftP); }else if(loc % 3 == 2){ // add one child swaps with left temparray = this.puzzle.clone(); temparray[loc] = temparray[loc - 1]; temparray[loc - 1] = 0; leftP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); leftP.setParent(this); children.add(leftP); } if(loc / 3 == 0){ //add one child swaps with lower temparray = this.puzzle.clone(); temparray[loc] = temparray[loc + 3]; temparray[loc + 3] = 0; downP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); downP.setParent(this); children.add(downP); }else if(loc / 3 == 1 ){ //add one child, swap with upper temparray = this.puzzle.clone(); temparray[loc] = temparray[loc - 3]; temparray[loc - 3] = 0; upP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); upP.setParent(this); children.add(upP); //add one child, swap with lower temparray = this.puzzle.clone(); temparray[loc] = temparray[loc + 3]; temparray[loc + 3] = 0; downP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); downP.setParent(this); children.add(downP); }else if (loc / 3 == 2 ){ //add one child, swap with upper temparray = this.puzzle.clone(); temparray[loc] = temparray[loc - 3]; temparray[loc - 3] = 0; upP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1); upP.setParent(this); children.add(upP); } return children; } public int h2(int[] list) // h2 = the sum of the distances of the tiles from their goal positions // for each item find its goal position // calculate how many positions it needs to move to get into that position { int gn = 0; int row = 0; int col = 0; for(int i = 0; i < list.length; i++) { if(list[i] != 0) { row = list[i] / 3; col = list[i] % 3; row = Math.abs(row - (i / 3)); col = Math.abs(col - (i % 3)); gn += row; gn += col; } } return gn; } public String toString() { String x = ""; for(int i = 0; i < this.puzzle.length; i++){ x += puzzle[i] + " "; if((i + 1) % 3 == 0) x += "\n"; } return x; } public int compareTo(Object input) { if (this.f_n < ((EightPuzzle) input).getF_n()) return -1; else if (this.f_n > ((EightPuzzle) input).getF_n()) return 1; return 0; } public boolean equals(EightPuzzle test){ if(this.f_n != test.getF_n()) return false; for(int i = 0 ; i < this.puzzle.length; i++) { if(this.puzzle[i] != test.puzzle[i]) return false; } return true; } public boolean mapEquals(EightPuzzle test){ for(int i = 0 ; i < this.puzzle.length; i++) { if(this.puzzle[i] != test.puzzle[i]) return false; } return true; } } proj1 import java.util.*; public class proj1 { /** * @param args */ public static void main(String[] args) { int[] p1d = {1, 4, 2, 3, 0, 5, 6, 7, 8}; int hueristic = 2; EightPuzzle start = new EightPuzzle(p1d, hueristic, 0); int[] win = { 0, 1, 2, 3, 4, 5, 6, 7, 8}; EightPuzzle goal = new EightPuzzle(win, hueristic, 0); astar(start, goal); } public static void astar(EightPuzzle start, EightPuzzle goal) { if(start.inversions() % 2 == 1) { System.out.println("Unsolvable"); return; } // function A*(start,goal) // closedset := the empty set // The set of nodes already evaluated. LinkedList<EightPuzzle> closedset = new LinkedList<EightPuzzle>(); // openset := set containing the initial node // The set of tentative nodes to be evaluated. priority queue PriorityQueue<EightPuzzle> openset = new PriorityQueue<EightPuzzle>(); openset.add(start); while(openset.size() > 0){ // x := the node in openset having the lowest f_score[] value EightPuzzle x = openset.peek(); // if x = goal if(x.mapEquals(goal)) { // return reconstruct_path(came_from, came_from[goal]) Stack<EightPuzzle> toDisplay = reconstruct(x); System.out.println("Printing solution... "); System.out.println(start.toString()); print(toDisplay); return; } // remove x from openset // add x to closedset closedset.add(openset.poll()); LinkedList <EightPuzzle> neighbor = x.getChildren(); // foreach y in neighbor_nodes(x) while(neighbor.size() > 0) { EightPuzzle y = neighbor.removeFirst(); // if y in closedset if(closedset.contains(y)){ // continue continue; } // tentative_g_score := g_score[x] + dist_between(x,y) // // if y not in openset if(!closedset.contains(y)){ // add y to openset openset.add(y); // } // } // } } public static void print(Stack<EightPuzzle> x) { while(!x.isEmpty()) { EightPuzzle temp = x.pop(); System.out.println(temp.toString()); } } public static Stack<EightPuzzle> reconstruct(EightPuzzle winner) { Stack<EightPuzzle> correctOutput = new Stack<EightPuzzle>(); while(winner.getParent() != null) { correctOutput.add(winner); winner = winner.getParent(); } return correctOutput; } }

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  • MySQL moving ibdata & ib_logfile

    - by XoR
    I'm trying to move ibdata & ib_logfile on ssd drive. I tried this way, but it don't work: service mysql stop cd /var/lib/ cp -ra mysql mysql_backup cp -a mysql/ibdata1 mysql/ib_logfile* /ssd_drive/mysql my.cnf looks like this (relevant parts): innodb_log_group_home_dir=/ssd_drive/mysql innodb_data_home_dir=/ssd_drive/mysql After all changes I get following errors: InnoDB: Unable to lock /ssd_drive/mysql/ibdata1, error: 13 InnoDB: Check that you do not already have another mysqld process Do I need to remove some lock files, or there is something else that I forgot... Also I setup mysql apparmor so it can rw on this directory, and rebooted afterward: /usr/sbin/mysqld { ................. /ssd_drive/mysql/* rw, ................. }

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  • Delphi label and asm weirdness?

    - by egon
    I written an asm function in Delphi 7 but it transforms my code to something else: function f(x: Cardinal): Cardinal; register; label err; asm not eax mov edx,eax shr edx, 1 and eax, edx bsf ecx, eax jz err mov eax, 1 shl eax, cl mov edx, eax add edx, edx or eax, edx ret err: xor eax, eax end; // compiled version f: push ebx // !!! not eax mov edx,eax shr edx, 1 and eax, edx bsf ecx, eax jz +$0e mov eax, 1 shl eax, cl mov edx, eax add edx, edx or eax, edx ret err: xor eax, eax mov eax, ebx // !!! pop ebx // !!! ret // the almost equivalent without asm function f(x: Cardinal): Cardinal; var c: Cardinal; begin x := not x; x := x and x shr 1; if x <> 0 then begin c := bsf(x); // bitscanforward x := 1 shl c; Result := x or (x shl 1) end else Result := 0; end; Why does it generate push ebx and pop ebx? And why does it do mov eax, ebx? It seems that it generates the partial stack frame because of the mov eax, ebx. This simple test generates mov eax, edx but doesn't generate that stack frame: function asmtest(x: Cardinal): Cardinal; register; label err; asm not eax and eax, 1 jz err ret err: xor eax, eax end; // compiled asmtest: not eax and eax, $01 jz +$01 ret xor eax, eax mov eax, edx // !!! ret It seems that it has something to do with the label err. If I remove that I don't get the mov eax, * part. Why does this happen? Made a bug report on Quality Central.

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  • Convert CRC-CCITT Kermit 16 DELPHI code to C#

    - by Mehdi Anis
    I am working on a function that will give me a Kermit CRC value from a HEX string. I have a piece of code in DELPHI. I am a .NET developer and need the code in C#. function CRC_16(cadena : string):word; var valuehex : word; i: integer; CRC : word; Begin CRC := 0; for i := 1 to length(cadena) do begin valuehex := ((ord(cadena[i]) XOR CRC) AND $0F) * $1081; CRC := CRC SHR 4; CRC := CRC XOR valuehex; valuehex := (((ord(cadena[i]) SHR 4) XOR LO(CRC)) AND $0F); CRC := CRC SHR 4; CRC := CRC XOR (valuehex * $1081); end; CRC_16 := (LO(CRC) SHL 8) OR HI(CRC); end; I got the code from this webpage: Kermit CRC in DELPHI I guess that Delphi function is correct. If any one can please convert the code to C# that will be great. I tried to convert to C#, but got lost in WORD data type and the LO function of Delphi. Thank you all.

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  • templete class c++

    - by inna karpasas
    hi! i try to design a templete for my universty project. i wrote the follwing cod: #ifndef _LinkedList_H_ #define _LinkedList_H_ #include "Link.h" #include <ostream> template <class L>//error one class LinkedList { private: Link<L> *pm_head; Link<L> * pm_tail; int m_numOfElements; Link<L>* FindLink(L * dataToFind); public: LinkedList(); ~LinkedList(); int GetNumOfElements(){return m_numOfElements;} bool Add( L * data); L *FindData(L * data); template <class L> friend ostream & operator<<(ostream& os,const LinkedList<L> listToprint);//error two L* GetDataOnTop(); bool RemoveFromHead(); L* Remove(L * toRemove); this templete uses the link class templete #ifndef _Link_H_ #define _Link_H_ template <class T>//error 3 class Link { private: T* m_data; Link* m_next; Link* m_prev; public: Link(T* data); ~Link(void); bool Link::operator ==(const Link& other)const; /*getters*/ Link* GetNext()const {return m_next;} Link* GetPrev()const {return m_prev;} T* GetData()const {return m_data;} //setters void SetNext(Link* next) {m_next = next;} void SetPrev(Link* prev) {m_prev = prev;} void SetData(T* data) {m_data = data;} }; error one: shadows template parm class L' error two:declaration ofclass L' error three: shadows template parm `class T' i dont understand what is the problem. i can relly usr your help thank you :)

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  • Converting "A* Search" code from C++ to Java [on hold]

    - by mr5
    Updated! I get this code from this site It's A* Search Algorithm(finding shortest path with heuristics) I modify most of variable names and some if conditions from the original version to satisfy my syntactic taste. It works in C++ (as I can't see any trouble with it) but fails in Java version. Java Code: String findPath(int startX, int startY, int finishX, int finishY) { @SuppressWarnings("unchecked") LinkedList<Node>[] nodeList = (LinkedList<Node>[]) new LinkedList<?>[2]; nodeList[0] = new LinkedList<Node>(); nodeList[1] = new LinkedList<Node>(); Node n0; Node m0; int nlIndex = 0; // queueList index // reset the node maps for(int y = 0;y < ROW_COUNT; ++y) { for(int x = 0;x < COL_COUNT; ++x) { close_nodes_map[y][x] = 0; open_nodes_map[y][x] = 0; } } // create the start node and push into list of open nodes n0 = new Node( startX, startY, 0, 0 ); n0.updatePriority( finishX, finishY ); nodeList[nlIndex].push( n0 ); open_nodes_map[startY][startX] = n0.getPriority(); // mark it on the open nodes map // A* search while( !nodeList[nlIndex].isEmpty() ) { LinkedList<Node> pq = nodeList[nlIndex]; // get the current node w/ the highest priority // from the list of open nodes n0 = new Node( pq.peek().getX(), pq.peek().getY(), pq.peek().getIterCount(), pq.peek().getPriority()); int x = n0.getX(); int y = n0.getY(); nodeList[nlIndex].pop(); // remove the node from the open list open_nodes_map[y][x] = 0; // mark it on the closed nodes map close_nodes_map[y][x] = 1; // quit searching when the goal state is reached //if((*n0).estimate(finishX, finishY) == 0) if( x == finishX && y == finishY ) { // generate the path from finish to start // by following the directions String path = ""; while( !( x == startX && y == startY) ) { int j = dir_map[y][x]; int c = '0' + ( j + Node.DIRECTION_COUNT / 2 ) % Node.DIRECTION_COUNT; path = (char)c + path; x += DIR_X[j]; y += DIR_Y[j]; } return path; } // generate moves (child nodes) in all possible directions for(int i = 0; i < Node.DIRECTION_COUNT; ++i) { int xdx = x + DIR_X[i]; int ydy = y + DIR_Y[i]; // boundary check if (!(xdx >= 0 && xdx < COL_COUNT && ydy >= 0 && ydy < ROW_COUNT)) continue; if ( ( gridMap.getData( ydy, xdx ) == GridMap.WALKABLE || gridMap.getData( ydy, xdx ) == GridMap.FINISH) && close_nodes_map[ydy][xdx] != 1 ) { // generate a child node m0 = new Node( xdx, ydy, n0.getIterCount(), n0.getPriority() ); m0.nextLevel( i ); m0.updatePriority( finishX, finishY ); // if it is not in the open list then add into that if( open_nodes_map[ydy][xdx] == 0 ) { open_nodes_map[ydy][xdx] = m0.getPriority(); nodeList[nlIndex].push( m0 ); // mark its parent node direction dir_map[ydy][xdx] = ( i + Node.DIRECTION_COUNT / 2 ) % Node.DIRECTION_COUNT; } else if( open_nodes_map[ydy][xdx] > m0.getPriority() ) { // update the priority info open_nodes_map[ydy][xdx] = m0.getPriority(); // update the parent direction info dir_map[ydy][xdx] = ( i + Node.DIRECTION_COUNT / 2 ) % Node.DIRECTION_COUNT; // replace the node // by emptying one queueList to the other one // except the node to be replaced will be ignored // and the new node will be pushed in instead while( !(nodeList[nlIndex].peek().getX() == xdx && nodeList[nlIndex].peek().getY() == ydy ) ) { nodeList[1 - nlIndex].push( nodeList[nlIndex].pop() ); } nodeList[nlIndex].pop(); // remove the wanted node // empty the larger size queueList to the smaller one if( nodeList[nlIndex].size() > nodeList[ 1 - nlIndex ].size() ) nlIndex = 1 - nlIndex; while( !nodeList[nlIndex].isEmpty() ) { nodeList[1 - nlIndex].push( nodeList[nlIndex].pop() ); } nlIndex = 1 - nlIndex; nodeList[nlIndex].push( m0 ); // add the better node instead } } } } return ""; // no route found } Output1: Legends . = PATH ? = START X = FINISH 3,2,1 = OBSTACLES (Misleading path) Output2: Changing these lines: n0 = new Node( a, b, c, d ); m0 = new Node( e, f, g, h ); to n0.set( a, b, c, d ); m0.set( e, f, g, h ); I get (I'm really confused) C++ Code: std::string A_Star::findPath(int startX, int startY, int finishX, int finishY) { typedef std::queue<Node> List_Container; List_Container nodeList[2]; // list of open (not-yet-tried) nodes Node n0; Node m0; int pqIndex = 0; // nodeList index // reset the node maps for(int y = 0;y < ROW_COUNT; ++y) { for(int x = 0;x < COL_COUNT; ++x) { close_nodes_map[y][x] = 0; open_nodes_map[y][x] = 0; } } // create the start node and push into list of open nodes n0 = Node( startX, startY, 0, 0 ); n0.updatePriority( finishX, finishY ); nodeList[pqIndex].push( n0 ); open_nodes_map[startY][startX] = n0.getPriority(); // mark it on the open nodes map // A* search while( !nodeList[pqIndex].empty() ) { List_Container &pq = nodeList[pqIndex]; // get the current node w/ the highest priority // from the list of open nodes n0 = Node( pq.front().getX(), pq.front().getY(), pq.front().getIterCount(), pq.front().getPriority()); int x = n0.getX(); int y = n0.getY(); nodeList[pqIndex].pop(); // remove the node from the open list open_nodes_map[y][x] = 0; // mark it on the closed nodes map close_nodes_map[y][x] = 1; // quit searching when the goal state is reached //if((*n0).estimate(finishX, finishY) == 0) if( x == finishX && y == finishY ) { // generate the path from finish to start // by following the directions std::string path = ""; while( !( x == startX && y == startY) ) { int j = dir_map[y][x]; char c = '0' + ( j + DIRECTION_COUNT / 2 ) % DIRECTION_COUNT; path = c + path; x += DIR_X[j]; y += DIR_Y[j]; } return path; } // generate moves (child nodes) in all possible directions for(int i = 0; i < DIRECTION_COUNT; ++i) { int xdx = x + DIR_X[i]; int ydy = y + DIR_Y[i]; // boundary check if (!( xdx >= 0 && xdx < COL_COUNT && ydy >= 0 && ydy < ROW_COUNT)) continue; if ( ( pGrid->getData(ydy,xdx) == WALKABLE || pGrid->getData(ydy, xdx) == FINISH) && close_nodes_map[ydy][xdx] != 1 ) { // generate a child node m0 = Node( xdx, ydy, n0.getIterCount(), n0.getPriority() ); m0.nextLevel( i ); m0.updatePriority( finishX, finishY ); // if it is not in the open list then add into that if( open_nodes_map[ydy][xdx] == 0 ) { open_nodes_map[ydy][xdx] = m0.getPriority(); nodeList[pqIndex].push( m0 ); // mark its parent node direction dir_map[ydy][xdx] = ( i + DIRECTION_COUNT / 2 ) % DIRECTION_COUNT; } else if( open_nodes_map[ydy][xdx] > m0.getPriority() ) { // update the priority info open_nodes_map[ydy][xdx] = m0.getPriority(); // update the parent direction info dir_map[ydy][xdx] = ( i + DIRECTION_COUNT / 2 ) % DIRECTION_COUNT; // replace the node // by emptying one nodeList to the other one // except the node to be replaced will be ignored // and the new node will be pushed in instead while ( !( nodeList[pqIndex].front().getX() == xdx && nodeList[pqIndex].front().getY() == ydy ) ) { nodeList[1 - pqIndex].push( nodeList[pqIndex].front() ); nodeList[pqIndex].pop(); } nodeList[pqIndex].pop(); // remove the wanted node // empty the larger size nodeList to the smaller one if( nodeList[pqIndex].size() > nodeList[ 1 - pqIndex ].size() ) pqIndex = 1 - pqIndex; while( !nodeList[pqIndex].empty() ) { nodeList[1-pqIndex].push(nodeList[pqIndex].front()); nodeList[pqIndex].pop(); } pqIndex = 1 - pqIndex; nodeList[pqIndex].push( m0 ); // add the better node instead } } } } return ""; // no route found } Output: Legends . = PATH ? = START X = FINISH 3,2,1 = OBSTACLES (Just right) From what I read about Java's documentation, I came up with the conclusion: C++'s std::queue<T>::front() == Java's LinkedList<T>.peek() Java's LinkedList<T>.pop() == C++'s std::queue<T>::front() + std::queue<T>::pop() What might I be missing in my Java version? In what way does it became different algorithmically from the C++ version?

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  • learning C++ from java, trying to make a linked list.

    - by kyeana
    I just started learning c++ (coming from java) and am having some serious problems with doing anything :P Currently, i am attempting to make a linked list, but must be doing something stupid cause i keep getting "void value not ignored as it ought to be" compile errors (i have it marked where it is throwing it bellow). If anyone could help me with what im doing wrong, i would be very grateful :) Also, I am not used to having the choice of passing by reference, address, or value, and memory management in general (currently i have all my nodes and the data declared on the heap). If anyone has any general advice for me, i also wouldn't complain :P Key code from LinkedListNode.cpp LinkedListNode::LinkedListNode() { //set next and prev to null pData=0; //data needs to be a pointer so we can set it to null for //for the tail and head. pNext=0; pPrev=0; } /* * Sets the 'next' pointer to the memory address of the inputed reference. */ void LinkedListNode::SetNext(LinkedListNode& _next) { pNext=&_next; } /* * Sets the 'prev' pointer to the memory address of the inputed reference. */ void LinkedListNode::SetPrev(LinkedListNode& _prev) { pPrev=&_prev; } //rest of class Key code from LinkedList.cpp #include "LinkedList.h" LinkedList::LinkedList() { // Set head and tail of linked list. pHead = new LinkedListNode(); pTail = new LinkedListNode(); /* * THIS IS WHERE THE ERRORS ARE. */ *pHead->SetNext(*pTail); *pTail->SetPrev(*pHead); } //rest of class

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  • unexpected output

    - by tech-ref
    hi, i wrote a function wich works as expected but i don't understand why the output is like that. function datatype prop = Atom of string | Not of prop | And of prop*prop | Or of prop*prop; (* XOR = (A And Not B) OR (Not A Or B) *) local fun do_xor (alpha,beta) = Or( And( alpha, Not(beta) ), Or(Not(alpha), beta)) in fun xor (alpha,beta) = do_xor(alpha,beta); end; test val result = xor(Atom "a",Atom "b"); output val result = Or (And (Atom #,Not #),Or (Not #,Atom #)) : prop thanks again (specially zeuxcg)

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  • How do i pass a number from a list as a parameter in scheme?

    - by wyatt
    I need to take a number from a list and convert it to a number so that i can pass it as a parameter. im trying to make a 1-bit adder in scheme. i've written the code for the or gate and the xor gate and also the half adder and now im trying to combine them all to make a full adder. im not sure if im going about it the right way. any input will be appreciated thank you. (define or-gate (lambda (a b) (if (= a 1) 1 (if (= b 1) 1 0)))) (define xor-gate (lambda (a b) (if (= a b) 0 1))) (define ha (lambda (a b) (list (xor-gate a b)(and-gate a b)))) (define fa (lambda (a b cin) (or-gate (cdr(ha cin (car (ha a b))))(cdr(ha a b))))) the issue i get when i run the program is that the half adder (ha) function outputs a list as a value and that makes the values incompatible with my other procedures because they require numbers and not lists. i feel like there is a simple solution but i dont know it.

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  • How to get better at solving Dynamic programming problems

    - by newbie
    I recently came across this question: "You are given a boolean expression consisting of a string of the symbols 'true', 'false', 'and', 'or', and 'xor'. Count the number of ways to parenthesize the expression such that it will evaluate to true. For example, there is only 1 way to parenthesize 'true and false xor true' such that it evaluates to true." I knew it is a dynamic programming problem so i tried to come up with a solution on my own which is as follows. Suppose we have a expression as A.B.C.....D where '.' represents any of the operations and, or, xor and the capital letters represent true or false. Lets say the number of ways for this expression of size K to produce a true is N. when a new boolean value E is added to this expression there are 2 ways to parenthesize this new expression 1. ((A.B.C.....D).E) ie. with all possible parenthesizations of A.B.C.....D we add E at the end. 2. (A.B.C.(D.E)) ie. evaluate D.E first and then find the number of ways this expression of size K can produce true. suppose T[K] is the number of ways the expression with size K produces true then T[k]=val1+val2+val3 where val1,val2,val3 are calculated as follows. 1)when E is grouped with D. i)It does not change the value of D ii)it inverses the value of D in the first case val1=T[K]=N.( As this reduces to the initial A.B.C....D expression ). In the second case re-evaluate dp[K] with value of D reversed and that is val1. 2)when E is grouped with the whole expression. //val2 contains the number of 'true' E will produce with expressions which gave 'true' among all parenthesized instances of A.B.C.......D i) if true.E = true then val2 = N ii) if true.E = false then val2 = 0 //val3 contains the number of 'true' E will produce with expressions which gave 'false' among all parenthesized instances of A.B.C.......D iii) if false.E=true then val3=( 2^(K-2) - N ) = M ie. number of ways the expression with size K produces a false [ 2^(K-2) is the number of ways to parenthesize an expression of size K ]. iv) if false.E=false then val3 = 0 This is the basic idea i had in mind but when i checked for its solution http://people.csail.mit.edu/bdean/6.046/dp/dp_9.swf the approach there was completely different. Can someone tell me what am I doing wrong and how can i get better at solving DP so that I can come up with solutions like the one given above myself. Thanks in advance.

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  • Linked List exercise, what am I doing wrong?

    - by Sean Ochoa
    Hey all. I'm doing a linked list exercise that involves dynamic memory allocation, pointers, classes, and exceptions. Would someone be willing to critique it and tell me what I did wrong and what I should have done better both with regards to style and to those subjects I listed above? /* Linked List exercise */ #include <iostream> #include <exception> #include <string> using namespace std; class node{ public: node * next; int * data; node(const int i){ data = new int; *data = i; } node& operator=(node n){ *data = *(n.data); } ~node(){ delete data; } }; class linkedList{ public: node * head; node * tail; int nodeCount; linkedList(){ head = NULL; tail = NULL; } ~linkedList(){ while (head){ node* t = head->next; delete head; if (t) head = t; } } void add(node * n){ if (!head) { head = n; head->next = NULL; tail = head; nodeCount = 0; }else { node * t = head; while (t->next) t = t->next; t->next = n; n->next = NULL; nodeCount++; } } node * operator[](const int &i){ if ((i >= 0) && (i < nodeCount)) throw new exception("ERROR: Invalid index on linked list.", -1); node *t = head; for (int x = i; x < nodeCount; x++) t = t->next; return t; } void print(){ if (!head) return; node * t = head; string collection; cout << "["; int c = 0; if (!t->next) cout << *(t->data); else while (t->next){ cout << *(t->data); c++; if (t->next) t = t->next; if (c < nodeCount) cout << ", "; } cout << "]" << endl; } }; int main (const int & argc, const char * argv[]){ try{ linkedList * myList = new linkedList; for (int x = 0; x < 10; x++) myList->add(new node(x)); myList->print(); }catch(exception &ex){ cout << ex.what() << endl; return -1; } return 0; }

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  • Installing Gtk2 on portable strawberry

    - by XoR
    I downloaded "strawberry-perl-5.12.2.0-portable" and "gtk+-bundle_2.22.1-20101227_win32". I extracted strawberry-perl in some directory and there I put gtk folder with gtk stuff. In portableshell.bat I changed Path env and added: "%drivep%\gtk\bin;%drivep%\gtk\lib;". Don't ask me why I added lib directory, I saw that some guy added it in some website. When I run in portableshell command: "pkg-config --libs --cflags gtk+-2.0" I get: c:\test>pkg-config --libs --cflags gtk+-2.0 -mms-bitfields -Ic:/test/gtk/include/gtk-2.0 -Ic:/test/gtk/lib/gtk-2.0/include - Ic:/test/gtk/include/atk-1.0 -Ic:/test/gtk/include/cairo -Ic:/test/gtk/include/g dk-pixbuf-2.0 -Ic:/test/gtk/include/pango-1.0 -Ic:/test/gtk/include/glib-2.0 -Ic :/test/gtk/lib/glib-2.0/include -Ic:/test/gtk/include -Ic:/test/gtk/include/free type2 -Ic:/test/gtk/include/libpng14 -Lc:/test/gtk/lib -lgtk-win32-2.0 -lgdk-wi n32-2.0 -latk-1.0 -lgio-2.0 -lpangowin32-1.0 -lgdi32 -lpangocairo-1.0 -lgdk_pixb uf-2.0 -lpango-1.0 -lcairo -lgobject-2.0 -lgmodule-2.0 -lgthread-2.0 -lglib-2.0 -lintl All folders looks fine, I also have complete log of compiling glib here. It looks like it doesn't compile because pkg-config gives bad data, or something. Does anyone have some idea how to make this thing work?

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  • Books and shellcode examples

    - by Xor
    i read "art of exploitation" and "gray hat hackers".Both these books examples written for x86 systems.i have a centrino laptop and an amd64 pc.I can't make work examples for stack based overflow.

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  • VHDL - Problem with std_logic_vector

    - by wretrOvian
    Hi, i'm coding a 4-bit binary adder with accumulator: library ieee; use ieee.std_logic_1164.all; entity binadder is port(n,clk,sh:in bit; x,y:inout std_logic_vector(3 downto 0); co:inout bit; done:out bit); end binadder; architecture binadder of binadder is signal state: integer range 0 to 3; signal sum,cin:bit; begin sum<= (x(0) xor y(0)) xor cin; co<= (x(0) and y(0)) or (y(0) and cin) or (x(0) and cin); process begin wait until clk='0'; case state is when 0=> if(n='1') then state<=1; end if; when 1|2|3=> if(sh='1') then x<= sum & x(3 downto 1); y<= y(0) & y(3 downto 1); cin<=co; end if; if(state=3) then state<=0; end if; end case; end process; done<='1' when state=3 else '0'; end binadder; The output : -- Compiling architecture binadder of binadder ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(15): No feasible entries for infix operator "xor". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(15): Type error resolving infix expression "xor" as type std.standard.bit. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): No feasible entries for infix operator "and". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in right operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): No feasible entries for infix operator "and". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in left operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in right operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Type error resolving infix expression "or" as type std.standard.bit. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(28): No feasible entries for infix operator "&". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(28): Type error resolving infix expression "&" as type ieee.std_logic_1164.std_logic_vector. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(39): VHDL Compiler exiting I believe i'm not handling std_logic_vector's correctly. Please tell me how? :(

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  • Unresolved External Symbol? error lnk2019

    - by jay
    Hello, I was wondering if anyone knew what this error meant. Thanks error LNK2019: unresolved external symbol "public: void __thiscall LinkedList,class std::allocator ::LocItem *::decreasekey(class PriorityList,class std::allocator ::LocItem * const &)" (?decreasekey@?$LinkedList@HPAVLocItem@?$PriorityList@HV?$basic_string@DU?$char_traits@D@std@@V?$allocator@D@2@@std@@@@@@QAEXABQAVLocItem@?$PriorityList@HV?$basic_string@DU?$char_traits@D@std@@V?$allocator@D@2@@std@@@@@Z) referenced in function "public: void __thiscall PriorityList,class std::allocator ::decreasekey(class PriorityList,class std::allocator ::Locator)" (?decreasekey@?$PriorityList@HV?$basic_string@DU?$char_traits@D@std@@V?$allocator@D@2@@std@@@@QAEXVLocator@1@@Z)

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  • Constraint Satisfaction Problem

    - by Carl Smotricz
    I'm struggling my way through Artificial Intelligence: A Modern Approach in order to alleviate my natural stupidity. In trying to solve some of the exercises, I've come up against the "Who Owns the Zebra" problem, Exercise 5.13 in Chapter 5. This has been a topic here on SO but the responses mostly addressed the question "how would you solve this if you had a free choice of problem solving software available?" I accept that Prolog is a very appropriate programming language for this kind of problem, and there are some fine packages available, e.g. in Python as shown by the top-ranked answer and also standalone. Alas, none of this is helping me "tough it out" in a way as outlined by the book. The book appears to suggest building a set of dual or perhaps global constraints, and then implementing some of the algorithms mentioned to find a solution. I'm having a lot of trouble coming up with a set of constraints suitable for modelling the problem. I'm studying this on my own so I don't have access to a professor or TA to get me over the hump - this is where I'm asking for your help. I see little similarity to the examples in the chapter. I was eager to build dual constraints and started out by creating (the logical equivalent of) 25 variables: nationality1, nationality2, nationality3, ... nationality5, pet1, pet2, pet3, ... pet5, drink1 ... drink5 and so on, where the number was indicative of the house's position. This is fine for building the unary constraints, e.g. The Norwegian lives in the first house: nationality1 = { :norway }. But most of the constraints are a combination of two such variables through a common house number, e.g. The Swede has a dog: nationality[n] = { :sweden } AND pet[n] = { :dog } where n can range from 1 to 5, obviously. Or stated another way: nationality1 = { :sweden } AND pet1 = { :dog } XOR nationality2 = { :sweden } AND pet2 = { :dog } XOR nationality3 = { :sweden } AND pet3 = { :dog } XOR nationality4 = { :sweden } AND pet4 = { :dog } XOR nationality5 = { :sweden } AND pet5 = { :dog } ...which has a decidedly different feel to it than the "list of tuples" advocated by the book: ( X1, X2, X3 = { val1, val2, val3 }, { val4, val5, val6 }, ... ) I'm not looking for a solution per se; I'm looking for a start on how to model this problem in a way that's compatible with the book's approach. Any help appreciated.

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  • Printf in assembler doesn't print

    - by Gaim
    Hi there, I have got a homework to hack program using buffer overflow ( with disassambling, program was written in C++, I haven't got the source code ). I have already managed it but I have a problem. I have to print some message on the screen, so I found out address of printf function, pushed address of "HACKED" and address of "%s" on the stack ( in this order ) and called that function. Called code passed well but nothing had been printed. I have tried to simulate the environment like in other place in the program but there has to be something wrong. Do you have any idea what I am doing wrong that I have no output, please? Thanks a lot EDIT: This program is running on Windows XP SP3 32b, written in C++, Intel asm there is the "hack" code CPU Disasm Address Hex dump Command Comments 0012F9A3 90 NOP ;hack begins 0012F9A4 90 NOP 0012F9A5 90 NOP 0012F9A6 89E5 MOV EBP,ESP 0012F9A8 83EC 7F SUB ESP,7F ;creating a place for working data 0012F9AB 83EC 7F SUB ESP,7F 0012F9AE 31C0 XOR EAX,EAX 0012F9B0 50 PUSH EAX 0012F9B1 50 PUSH EAX 0012F9B2 50 PUSH EAX 0012F9B3 89E8 MOV EAX,EBP 0012F9B5 83E8 09 SUB EAX,9 0012F9B8 BA 1406EDFF MOV EDX,FFED0614 ;address to jump, it is negative because there mustn't be 00 bytes 0012F9BD F7DA NOT EDX 0012F9BF FFE2 JMP EDX ;I have to jump because there are some values overwritten by the program 0012F9C1 90 NOP 0012F9C2 0090 00000000 ADD BYTE PTR DS:[EAX],DL 0012F9C8 90 NOP 0012F9C9 90 NOP 0012F9CA 90 NOP 0012F9CB 90 NOP 0012F9CC 6C INS BYTE PTR ES:[EDI],DX ; I/O command 0012F9CD 65:6E OUTS DX,BYTE PTR GS:[ESI] ; I/O command 0012F9CF 67:74 68 JE SHORT 0012FA3A ; Superfluous address size prefix 0012F9D2 2069 73 AND BYTE PTR DS:[ECX+73],CH 0012F9D5 203439 AND BYTE PTR DS:[EDI+ECX],DH 0012F9D8 34 2C XOR AL,2C 0012F9DA 2066 69 AND BYTE PTR DS:[ESI+69],AH 0012F9DD 72 73 JB SHORT 0012FA52 0012F9DF 74 20 JE SHORT 0012FA01 0012F9E1 3120 XOR DWORD PTR DS:[EAX],ESP 0012F9E3 6C INS BYTE PTR ES:[EDI],DX ; I/O command 0012F9E4 696E 65 7300909 IMUL EBP,DWORD PTR DS:[ESI+65],-6F6FFF8D 0012F9EB 90 NOP 0012F9EC 90 NOP 0012F9ED 90 NOP 0012F9EE 31DB XOR EBX,EBX ; hack continues 0012F9F0 8818 MOV BYTE PTR DS:[EAX],BL ; writing 00 behind word "HACKED" 0012F9F2 83E8 06 SUB EAX,6 0012F9F5 50 PUSH EAX ; address of "HACKED" 0012F9F6 B8 3B8CBEFF MOV EAX,FFBE8C3B 0012F9FB F7D0 NOT EAX 0012F9FD 50 PUSH EAX ; address of "%s" 0012F9FE B8 FFE4BFFF MOV EAX,FFBFE4FF 0012FA03 F7D0 NOT EAX 0012FA05 FFD0 CALL EAX ;address of printf This code is really ugly because I am new in assembler and there mustn't be null bytes because of buffer-overflow bug

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  • Java - Collections.sort() performance

    - by msr
    Hello, Im using Collections.sort() to sort a LinkedList whose elements implements Comparable interface, so they are sorted in a natural order. In the javadoc documentation its said this method uses mergesort algorithm wich has n*log(n) performance. My question is if there is a more efficient algorithm to sort my LinkedList? The size of that list could be very high and sort will be also very frequent. Thanks!

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