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  • Recursion in assembly?

    - by Davis
    I'm trying to get a better grasp of assembly, and I am a little confused about how to recursively call functions when I have to deal with registers, popping/pushing, etc. I am embedding x86 assembly in C++. Here I am trying to make a method which given an array of integers will build a linked list containing these integers in the order they appear in the array. I am doing this by calling a recursive function: insertElem (struct elem *head, struct elem *newElem, int data) -head: head of the list -data: the number that will be inserted at the end of a list -newElem: points to the location in memory where I will store the new element (data field) My problem is that I keep overwriting the registers instead of a typical linked list. For example, if I give it an array {2,3,1,8,3,9} my linked-list will return the first element (head) and only the last element, because the elements keep overwriting each other after head is no longer null. So here my linked list looks something like: 2--9 instead of 2--3--1--8--3--9 I feel like I don't have a grasp on how to organize and handle the registers. newElem is in EBX and just keeps getting rewritten. Thanks in advance!

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  • Recursion Interview Questions [closed]

    - by halivingston
    Given a string, "ABC", print all permutations Given a dollar bill, fill out possible ways it can summed up using .25, .10, .5, etc. Given a phone number (123-456), print out all it's word counter parters like (ADG-XYZ) A B C D E F G H I J K L M N O P In the above 2D matrix, print all possible words (just literally all words, and sure we could check if it's exists in a dictionary). The base case is I think here is that reaching the same i, j positions. Any others you can think of?

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  • Recursion through a directory tree in PHP

    - by phphelpplz
    I have a set of folders that has a depth of at least 4 or 5 levels. I'm looking to recurse through the directory tree as deep as it goes, and iterate over every file. I've gotten the code to go down into the first sets of subdirectories, but no deeper, and I'm not sure why. Any ideas? $count = 0; $dir = "/Applications/MAMP/htdocs/idahohotsprings.com"; function recurseDirs($main, $count){ $dir = "/Applications/MAMP/htdocs/idahohotsprings.com"; $dirHandle = opendir($main); echo "here"; while($file = readdir($dirHandle)){ if(is_dir($file) && $file != '.' && $file != '..'){ echo "isdir"; recurseDirs($file); } else{ $count++; echo "$count: filename: $file in $dir/$main \n<br />"; } } } recurseDirs($dir, $count); ?

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  • Python recursion with list returns None

    - by newman
    def foo(a): a.append(1) if len(a) > 10: print a return a else: foo(a) Why this recursive function returns None (see transcript below)? I can't quite understand what I am doing wrong. In [263]: x = [] In [264]: y = foo(x) [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] In [265]: print y None

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  • ruby inject recursion?

    - by Matt Humphrey
    the goal is to start with ['a','b','c'] and end up with {'a'={'b'={'c'={}}}} so, getting my bearings, i did this: ruby-1.8.7-p174 ['a','b','c'].inject({}){|h,v| h.update(v = {})} = {"a"={}, "b"={}, "c"={}} and then figured, if i actually pass on the result hash, it will recurse and nest, but: ruby-1.8.7-p174 ['a','b','c'].inject({}){|h,v| h.update(v = {}); h[v]} = {} why is this? any idea how to achieve the desired result in an elegant one-liner?

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  • Javascript Recursion

    - by rpophessagr
    I have an ajax call and would like to recall it once I finish parsing and animating the result into the page. And that's where I'm getting stuck. I was able to recall the function, but it seems to not take into account the delays in the animation. i.e. The console keeps outputting the values at a wild pace. I thought setInterval might help with the interval being the sum of the length of my delays, but I can't get that to work... function loadEm(){ var result=new Array(); $.getJSON("jsonCall.php",function(results){ $.each(results, function(i, res){ rand = (Math.floor(Math.random()*11)*1000)+2000; fullRand += rand; console.log(fullRand); $("tr:first").delay(rand).queue(function(next) { doStuff(res); next(); }); }); var int=self.setInterval("loadEm()",fullRand); }); } });

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  • Java Recursion Triangle Standing on Tip

    - by user1629075
    I was wondering how to create triangle out of asterisks on its tip rather than on its base. I have the code for making it stand on its base: public static String printTriangle (int count) { if( count <= 0 ) return ""; String p = printTriangle(count - 1); p = p + "*"; System.out.print(p); System.out.print("\n"); return p; } But then I'm stuck on how to have the greatest number of stars on the top, and then the next least, and so on. I was thinking something along the terms of having (count - p) to have the input of rows be subtracted from the amount of decrease, but then i was confused by this idea because p is string. EDIT: I tried changing the position of printTriangle(count - 1) using my original method without iterations and got 1 star per each line; how can I fix this? public class triangles { public static void main(String[] args) { printTriangle(5); } public static String printTriangle (int count) { if( count <= 0 ) return ""; String p = ""; p = p + "*"; System.out.print(p); System.out.print("\n"); p = printTriangle(count - 1); return p; } }

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  • Python list recursion type error

    - by Jacob J Callahan
    I can't seem to figure out why the following code is giving me a TypeError: 'type' object is not iterable pastebin: http://pastebin.com/VFZYY4v0 def genList(self): #recursively generates a sorted list of child node values numList = [] if self.leftChild != 'none': numList.extend(self.leftChild.genList()) #error numList.extend(list((self.Value,))) if self.rightChild != 'none': numList.extend(self.rightChild.genList()) #error return numList code that adds child nodes (works correctly) def addChild(self, child): #add a child node. working if child.Value < self.Value: if self.leftChild == 'none': self.leftChild = child child.parent = self else: self.leftChild.addChild(child) elif child.Value > self.Value: if self.rightChild == 'none': self.rightChild = child child.parent = self else: self.rightChild.addChild(child) Any help would be appreciated. Full interpreter session: >>> import BinTreeNode as BTN >>> node1 = BTN.BinaryTreeNode(5) >>> node2 = BTN.BinaryTreeNode(2) >>> node3 = BTN.BinaryTreeNode(12) >>> node3 = BTN.BinaryTreeNode(16) >>> node4 = BTN.BinaryTreeNode(4) >>> node5 = BTN.BinaryTreeNode(13) >>> node1.addChild(node2) >>> node1.addChild(node3) >>> node1.addChild(node4) >>> node1.addChild(node5) >>> node4.genList() <class 'list'> >>> node1.genList() Traceback (most recent call last): File "<interactive input>", line 1, in <module> File "C:...\python\BinTreeNode.py", line 47, in genList numList.extend(self.leftChild.genList()) #error File "C:...\python\BinTreeNode.py", line 52, in genList TypeError: 'type' object is not iterable

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  • Recursion: using values passed in parameters

    - by Tom Lilletveit
    I got this line of code that pops a int of the array saves it to int element then removes it from array. then in the return statement return CountCriticalVotes(rest, blockIndex + element); it ads it to the blockIndex variable and if it reaches 10 before the array is empty it returns 1. But my problem is this, I do not want it to add up all the values in the array in the parameter, but only add one then revert the parameter value back to it´s original state, then add a new, revert etc... How would i do this? int NumCriticalVotes :: CountCriticalVotes(Vector<int> & blocks, int blockIndex) { if (blockIndex >= 10) { return 1; } if (blocks.isEmpty()) { return 0; } else { int element = blocks.get(0); Vector<int> rest = blocks; rest.remove(0); return CountCriticalVotes(rest, blockIndex + element);

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  • Removing Left Recursion in ANTLR

    - by prosseek
    As is explained in http://stackoverflow.com/questions/2652060/removing-left-recursion , there are two ways to remove the left recursion. Modify the original grammar to remove the left recursion using some procedure Write the grammar originally not to have the left recursion What people normally use for removing (not having) the left recursion with ANTLR? I've used flex/bison for parser, but I need to use ANTLR. The only thing I'm concerned about using ANTLR (or LL parser in genearal) is left recursion removal. In practical sense, how serious of removing left recursion in ANTLR? Is this a showstopper in using ANTLR? Or, nobody cares about it in ANTLR community? I like the idea of AST generation of ANTLR. In terms of getting AST quick and easy way, which method (out of the 2 removing left recursion methods) is preferable?

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  • Is recursion really bad?

    - by dotneteer
    After my previous post about the stack space, it appears that there is perception from the feedback that recursion is bad and we should avoid deep recursion. After writing a compiler, I know that the modern computer and compiler are complex enough and one cannot automatically assume that a hand crafted code would out-perform the compiler optimization. The only way is to do some prototype to find out. So why recursive code may not perform as well? Compilers place frames on a stack. In additional to arguments and local variables, compiles also need to place frame and program pointers on the frame, resulting in overheads. So why hand-crafted code may not performance as well? The stack used by a compiler is a simpler data structure and can grow and shrink cleanly. To replace recursion with out own stack, our stack is allocated in the heap that is far more complicated to manage. There could be overhead as well if the compiler needs to mark objects for garbage collection. Compiler also needs to worry about the memory fragmentation. Then there is additional complexity: CPUs have registers and multiple levels of cache. Register access is a few times faster than in-CPU cache access and is a few 10s times than on-board memory access. So it is up to the OS and compiler to maximize the use of register and in-CPU cache. For my particular problem, I did an experiment to rewrite my c# version of recursive code with a loop and stack approach. So here are the outcomes of the two approaches:   Recursive call Loop and Stack Lines of code for the algorithm 17 46 Speed Baseline 3% faster Readability Clean Far more complex So at the end, I was able to achieve 3% better performance with other drawbacks. My message is never assuming your sophisticated approach would automatically work out better than a simpler approach with a modern computer and compiler. Gage carefully before committing to a more complex approach.

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  • Stack Trace Logger [migrated]

    - by Chris Okyen
    I need to write a parent Java class that classes using recursion can extend. The parent class will be be able to realize whenever the call stack changes ( you enter a method, temporarily leave it to go to another method call, or you are are finsihed with the method ) and then print it out. I want it to print on the console, but clear the console as well every time so it shows the stack horizantaly so you can see the height of each stack to see what popped off and what popped on... Also print out if a baseline was reached for recursive functions. First. How can I using the StackTraceElements and Thread classes to detect automatically whenever the stack has popped or pushed an element on without calling it manually? Second, how would I do the clearing thing? For instance , if I had the code: public class recursion(int i) { private static void recursion(int i) { if( i < 10) System.out.println('A'); else { recursion(i / 10 ); System.out.println('B'); } } public static void main(String[] argv) { recursion(102); } } It would need to print out the stack when entering main(), when entering recursion(102) from main(), when it enters recursion(102 / 10), which is recursion(10), from recursion(102), when it enters recursion(10 / 10), which is recursion(1) from recursion(10). Print out a message out when it reaches the baseline recursion(1).. then print out the stacks of reversed revisitation of function recursion(10), recursion(102) and main(). finally print out we are exiting main().

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  • Grub2 mutual dependency issue

    - by A T
    For various reasons I am installing .deb dependencies for grub2 using dpkg directly (rather than apt-get). root@ubuntu:/dl# dpkg -i grub-gfxpayload-lists_0.6_amd64.deb Selecting previously unselected package grub-gfxpayload-lists. (Reading database ... 249808 files and directories currently installed.) Preparing to unpack grub-gfxpayload-lists_0.6_amd64.deb ... Unpacking grub-gfxpayload-lists (0.6) ... dpkg: dependency problems prevent configuration of grub-gfxpayload-lists: grub-gfxpayload-lists depends on grub-pc (>= 1.99~20101210-1ubuntu2); however: Package grub-pc is not configured yet. dpkg: error processing package grub-gfxpayload-lists (--install): dependency problems - leaving unconfigured Processing triggers for man-db (2.6.7.1-1) ... Errors were encountered while processing: grub-gfxpayload-lists By configure I assume it means install+configure, so I tried: root@ubuntu:/dl# dpkg -i grub-pc_2.02~beta2-9_amd64.deb (Reading database ... 249818 files and directories currently installed.) Preparing to unpack grub-pc_2.02~beta2-9_amd64.deb ... Unpacking grub-pc (2.02~beta2-9) over (2.02~beta2-9) ... dpkg: dependency problems prevent configuration of grub-pc: grub-pc depends on grub2-common (= 2.02~beta2-9); however: Package grub2-common is not installed. grub-pc depends on grub-pc-bin (= 2.02~beta2-9); however: Package grub-pc-bin is not installed. grub-pc depends on grub-gfxpayload-lists; however: Package grub-gfxpayload-lists is not configured yet. dpkg: error processing package grub-pc (--install): dependency problems - leaving unconfigured Processing triggers for man-db (2.6.7.1-1) ... Errors were encountered while processing: grub-pc How do I solve this problem?

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  • Why is the recursion idiom in Haskell "'n+1' and 'n'" and not "'n' and 'n-1'"?

    - by rulfzid
    I'm working my way through Graham Hutton's Haskell book, and in his recursion chapter, he often pattern-matches on "n+1", as in: myReplicate1 0 _ = [] myReplicate1 (n+1) x = x : myReplicate1 n x Why that and not the following, which (1) seems functionally identical and (2) more intuitive in terms of understanding what's happening with the recursion: myReplicate2 0 _ = [] myReplicate2 n x = x : myReplicate2 (n-1) x Is there something I'm missing here? Or is it just a matter of style?

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  • Need help understanding a recursion example in Python

    - by Ali Mustafa
    Python is my first programming language, and I'm learning it from "How to Think Like a Computer Scientist". In Chapter 5 the author gives the following example on recursion: def factorial(n): if n == 0: return 1 else: recurse = factorial(n-1) result = n * recurse return result I understand that if n = 3, then the function will execute the second branch. But what I don't understand is what happens when the function enters the second branch. Can someone explain it to me?

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  • DNS "recursion not available" using a Cisco AnyConnect VPN connection

    - by codeape
    Does anyone have experience with configuring Cisco AnyConnect VPN? We have a problem with client DNS name resolution when connected over VPN. To me, it looks as if the Cisco AnyConnect VPN client intercepts DNS queries from the clients. Can someone confirm that the AnyConnect VPN client in fact does this (intercepts DNS traffic)? Where is this configured on the VPN server? EDIT: Here's how the routing table changes when I connect to the VPN: [~] $ diff -u /tmp/route_normal /tmp/route_vpn --- /tmp/route_normal 2010-01-20 19:23:47.000000000 +0100 +++ /tmp/route_vpn 2010-01-20 19:24:46.000000000 +0100 @@ -1,6 +1,10 @@ Kernel IP routing table Destination Gateway Genmask Flags Metric Ref Use Iface +xxx.xxx.xx.xx.i 10.0.0.1 255.255.255.255 UGH 0 0 0 ath0 172.16.53.0 * 255.255.255.0 U 0 0 0 vmnet1 10.0.0.0 * 255.255.255.0 U 0 0 0 ath0 +172.17.20.0 * 255.255.255.0 U 0 0 0 cscotun 0 +192.168.111.0 172.17.20.212 255.255.255.0 UG 0 0 0 cscotun 0 172.16.140.0 * 255.255.255.0 U 0 0 0 vmnet8 +172.16.0.0 172.17.20.212 255.255.0.0 UG 0 0 0 cscotun 0 default 10.0.0.1 0.0.0.0 UG 0 0 0 ath0 EDIT 2: The IT guy has done "something" on the VPN endpoint. Now I get "recursion not available" when doing nslookup. The DNS servers have recursion enabled. So it must be the Cisco VPN DNS interception messing this up. ubuntu@domU-12-31-39-00-ED-14:~$ /opt/cisco/vpn/bin/vpn connect xxx.xxxxxx.xx ... >> Please enter your username and password ... >> notice: Establishing VPN... >> state: Connected >> notice: VPN session established to ... ubuntu@domU-12-31-39-00-ED-14:~$ nslookup www.vg.no ;; Got recursion not available from ..., trying next server ;; Got recursion not available from ..., trying next server ;; Got recursion not available from ..., trying next server ;; Got recursion not available from ..., trying next server Server: 172.16.0.23 Address: 172.16.0.23#53 ** server can't find www.vg.no.compute-1.internal: REFUSED ubuntu@domU-12-31-39-00-ED-14:~$ ping 195.88.55.16 PING 195.88.55.16 (195.88.55.16) 56(84) bytes of data. 64 bytes from 195.88.55.16: icmp_seq=1 ttl=240 time=110 ms 64 bytes from 195.88.55.16: icmp_seq=2 ttl=240 time=111 ms 64 bytes from 195.88.55.16: icmp_seq=3 ttl=240 time=109 ms ^C --- 195.88.55.16 ping statistics --- 3 packets transmitted, 3 received, 0% packet loss, time 2017ms rtt min/avg/max/mdev = 109.953/110.379/111.075/0.496 ms

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  • PCI compliance - Setting BIND to no recursion, cURL can't access external sites

    - by Exit
    I was running a PCI scan and was following direction to change the BIND options from: // recursion no; allow-recursion { trusted;}; allow-notify { trusted;}; allow-transfer { trusted;}; to: recursion no; allow-recursion { none;}; allow-notify { trusted;}; allow-transfer { none;}; The end result was that cURL operations stopped being able to access external sites. I realize that not everything will be 100% for PCI compliance, but can someone explain if there is a way to balance this for both PCI compliance and function?

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  • Loops, Recursion and Memoization in JavaScript

    - by Ken Dason
    Originally posted on: http://geekswithblogs.net/kdason/archive/2013/07/25/loops-recursion-and-memoization-in-javascript.aspxAccording to Wikipedia, the factorial of a positive integer n (denoted by n!) is the product of all positive integers less than or equal to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120. The value of 0! is 1. We can use factorials to demonstrate iterative loops and recursive functions in JavaScript.  Here is a function that computes the factorial using a for loop: Output: Time Taken: 51 ms Here is the factorial function coded to be called recursively: Output: Time Taken: 165 ms We can speed up the recursive function with the use of memoization.  Hence,  if the value has previously been computed, it is simply returned and the recursive call ends. Output: Time Taken: 17 ms

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  • Understanding Backtracking in C++

    - by nikhil
    I have a good basic understanding of the fundamentals of C++, I also have an understanding of how recursion works too. I came across certain problems like the classic eight queens problem and solving a Sudoku with Backtracking. I realize that I'm quite lost when it comes to this, I can't seem to be able to get my mind around the concept of going back in the recursion stack and starting again in order to solve the problem. It seems easy with a pen and paper but when it comes to writing code for this, I'm confused on how to begin attacking these problems. It would be helpful if there were a tutorial aimed at beginners to backtracking or if there were a good book where this was covered. If somebody can shed light on this topic or give me some links to decent references, I'd be really grateful. And yes I do know that it would be easier in functional languages but I'd like to understand the implementation in imperative languages too.

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