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  • DataContractJsonSerializer produces list of hashes instead of hash

    - by Jacques
    I would expect a Dictionary object of the form: Dictionary<string,string> dict = new Dictionary<string,string>() {["blah", "bob"], ["blahagain", "bob"]}; to serialize into JSON in the form of: { "blah": "bob", "blahagain": "bob" } NOT [ { "key": "blah", "value": "bob" }, { "key": "blahagain", "value": "bob"}] What is the reason for what appears to be a monstrosity of a generic attempt at serializing collections? The DataContractJsonSerializer uses the ISerializable interface to produce this thing. It seems to me as though somebody has taken the XML output from ISerializable and mangled this thing out of it. Is there a way to override the default serialization used by .Net here? Could I just derive from Dictionary and override the Serialization methods? Posting to hear of any caveats or suggestions people might have.

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  • Visual basic 6.0 - ComputeHash invalid procedure call or argument error

    - by Mohan Babu Vijaya Gopal
    I am getting the error "invalid procedure call or arguments" at the step computeHash(). Any help highly appreciated. Private Sub Form_Load() Dim rngcsp As New RNGCryptoServiceProvider '= new RNGCryptoServiceProvider() Dim u8 As Encoding 'u8 = Encoding.UTF8 Dim minSaltSize As Integer Dim maxSaltSize As Integer Dim saltSize As Integer minSaltSize = 4 maxSaltSize = 8 Dim randm As Random Set randm = New Random Dim saltBytes() As Byte ReDim saltBytes(saltSize) Set rngcsp = New RNGCryptoServiceProvider rngcsp.GetNonZeroBytes (saltBytes) Dim plainTextBytes() As Byte plainTextBytes() = ConvertStringToUtf8Bytes("Mohan") Dim plainTextBytesLen As Long plainTextBytesLen = UBound(plainTextBytes) - LBound(plainTextBytes) + 1 Dim saltBytesLen As Long saltBytesLen = UBound(saltBytes) - LBound(saltBytes) + 1 Dim plainTextWithSaltBytes() As Byte ReDim plainTextWithSaltBytes(plainTextBytesLen + saltBytesLen) For i = 0 To plainTextBytesLen - 1 plainTextWithSaltBytes(i) = plainTextBytes(i) Next For i = 0 To saltBytesLen - 1 plainTextWithSaltBytes(i) = saltBytes(i) Next 'Dim hash As HashAlgorithm = New MD5CryptoServiceProvider() Dim hash12 As New SHA256Managed 'SHA256Managed Dim totLen As Integer totLen = plainTextBytesLen + saltBytesLen Dim str As String Dim hashBytes() As Byte 'With hashBytes = hash12.computeHash(plainTextWithSaltBytes) ', 0, totLen) 'End With End Sub

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  • How to assign the array key a value, when the key name is itself a variable

    - by Matrym
    How do I identify an item in a hash array if the key of the array is only known within a variable? For example: var key = "myKey"; var array = {myKey: 1, anotherKey: 2}; alert(array.key); Also, how would I assign a value to that key, having identified it with the variable? This is, of course, assuming that I must use the variable key to identify which item in the array to alert. Thanks in advance!

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  • Getting Facebook email/email hash.

    - by Zenzen
    I'll make it short: is it possible to get the user's facebook email adress (or a hashed email adress), so I can let's say compare his facebook email with his email in my database? I'm trying to get a FacebookUser and then use facebookUser.email_hash but that returns nothing.

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  • Undefined reference to cmph functions even after installing cpmh library

    - by user1242145
    I am using gcc 4.4.3 on ubuntu. I installed cmph library tools 0.9-1 using command sudo apt-get install libcmph-tools Now, when I tried to compile example program vector_adapter_ex1.c , gcc is able to detect cmph.h library in its include file but is showing multiple errors like vector_adapter_ex1.c:(.text+0x93): undefined reference to cmph_io_vector_adapter' vector_adapter_ex1.c:(.text+0xa3): undefined reference tocmph_config_new' vector_adapter_ex1.c:(.text+0xbb): undefined reference to cmph_config_set_algo' vector_adapter_ex1.c:(.text+0xcf): undefined reference tocmph_config_set_mphf_fd' even though, these are all defined in the source code of the cmph library. Could anyone tell the error that might have occurred or suggest an alternate method to go about building minimal perfect hash functions.

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  • Sorting in Hash Maps in Java

    - by Crystal
    I'm trying to get familiar with Collections. I have a String which is my key, email address, and a Person object (firstName, lastName, telephone, email). I read in the Java collections chapter on Sun's webpages that if you had a HashMap and wanted it sorted, you could use a TreeMap. How does this sort work? Is it based on the compareTo() method you have in your Person class? I overrode the compareTo() method in my Person class to sort by lastName. But it isn't working properly and was wondering if I have the right idea or not. getSortedListByLastName at the bottom of this code is where I try to convert to a TreeMap. Also, if this is the correct way to do it, or one of the correct ways to do it, how do I then sort by firstName since my compareTo() is comparing by lastName. import java.util.*; public class OrganizeThis { /** Add a person to the organizer @param p A person object */ public void add(Person p) { staff.put(p.getEmail(), p); //System.out.println("Person " + p + "added"); } /** * Remove a Person from the organizer. * * @param email The email of the person to be removed. */ public void remove(String email) { staff.remove(email); } /** * Remove all contacts from the organizer. * */ public void empty() { staff.clear(); } /** * Find the person stored in the organizer with the email address. * Note, each person will have a unique email address. * * @param email The person email address you are looking for. * */ public Person findByEmail(String email) { Person aPerson = staff.get(email); return aPerson; } /** * Find all persons stored in the organizer with the same last name. * Note, there can be multiple persons with the same last name. * * @param lastName The last name of the persons your are looking for. * */ public Person[] find(String lastName) { ArrayList<Person> names = new ArrayList<Person>(); for (Person s : staff.values()) { if (s.getLastName() == lastName) { names.add(s); } } // Convert ArrayList back to Array Person nameArray[] = new Person[names.size()]; names.toArray(nameArray); return nameArray; } /** * Return all the contact from the orgnizer in * an array sorted by last name. * * @return An array of Person objects. * */ public Person[] getSortedListByLastName() { Map<String, Person> sorted = new TreeMap<String, Person>(staff); ArrayList<Person> sortedArrayList = new ArrayList<Person>(); for (Person s: sorted.values()) { sortedArrayList.add(s); } Person sortedArray[] = new Person[sortedArrayList.size()]; sortedArrayList.toArray(sortedArray); return sortedArray; } private Map<String, Person> staff = new HashMap<String, Person>(); public static void main(String[] args) { OrganizeThis testObj = new OrganizeThis(); Person person1 = new Person("J", "W", "111-222-3333", "[email protected]"); Person person2 = new Person("K", "W", "345-678-9999", "[email protected]"); Person person3 = new Person("Phoebe", "Wang", "322-111-3333", "[email protected]"); Person person4 = new Person("Nermal", "Johnson", "322-342-5555", "[email protected]"); Person person5 = new Person("Apple", "Banana", "123-456-1111", "[email protected]"); testObj.add(person1); testObj.add(person2); testObj.add(person3); testObj.add(person4); testObj.add(person5); System.out.println(testObj.findByEmail("[email protected]")); System.out.println("------------" + '\n'); Person a[] = testObj.find("W"); for (Person p : a) System.out.println(p); System.out.println("------------" + '\n'); a = testObj.find("W"); for (Person p : a) System.out.println(p); System.out.println("SORTED" + '\n'); a = testObj.getSortedListByLastName(); for (Person b : a) { System.out.println(b); } } } Person class: public class Person implements Comparable { String firstName; String lastName; String telephone; String email; public Person() { firstName = ""; lastName = ""; telephone = ""; email = ""; } public Person(String firstName) { this.firstName = firstName; } public Person(String firstName, String lastName, String telephone, String email) { this.firstName = firstName; this.lastName = lastName; this.telephone = telephone; this.email = email; } public String getFirstName() { return firstName; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getLastName() { return lastName; } public void setLastName(String lastName) { this.lastName = lastName; } public String getTelephone() { return telephone; } public void setTelephone(String telephone) { this.telephone = telephone; } public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } public int compareTo(Object o) { String s1 = this.lastName + this.firstName; String s2 = ((Person) o).lastName + ((Person) o).firstName; return s1.compareTo(s2); } public boolean equals(Object otherObject) { // a quick test to see if the objects are identical if (this == otherObject) { return true; } // must return false if the explicit parameter is null if (otherObject == null) { return false; } if (!(otherObject instanceof Person)) { return false; } Person other = (Person) otherObject; return firstName.equals(other.firstName) && lastName.equals(other.lastName) && telephone.equals(other.telephone) && email.equals(other.email); } public int hashCode() { return this.email.toLowerCase().hashCode(); } public String toString() { return getClass().getName() + "[firstName = " + firstName + '\n' + "lastName = " + lastName + '\n' + "telephone = " + telephone + '\n' + "email = " + email + "]"; } }

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  • Is it possible to get identical SHA1 hash?

    - by drozzy
    Given two different strings S1 and S2 (S1 != S2) is it possible that: SHA1(S1) == SHA1(S2) is True? If yes - with what probability? Is there a upper bound on the length of a string, for which probably of getting duplicates is 0? If not - why not? Thanks

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  • Is it possible to convert a 40-character SHA1 hash to a 20-character SHA1 hash?

    - by ewitch
    My problem is a bit hairy, and I may be asking the wrong questions, so please bear with me... I have a legacy MySQL database which stores the user passwords & salts for a membership system. Both of these values have been hashed using the Ruby framework - roughly like this: hashedsalt = Digest::SHA1.hexdigest("--#{Time.now.to_s}--#{login}--") hashedpassword = Digest::SHA1.hexdigest("#{hashedsalt}:#{password}") So both values are stored as 40-character strings (varchar(40)) in MySQL. Now I need to import all of these users into the ASP.NET membership framework for a new web site, which uses a SQL Server database. It is my understanding that the the way I have ASP.NET membership configured, the user passwords and salts are also stored in the membership database (in table aspnet_Membership) as SHA1 hashes, which are then Base64 encoded (see here for details) and stored as nvarchar(128) data. But from the length of the Base64 encoded strings that are stored (28 characters) it seems that the SHA1 hashes that ASP.NET membership generates are only 20 characters long, rather than 40. From some other reading I have been doing I am thinking this has to do with the number of bits per character/character set/encoding or something related. So is there some way to convert the 40-character SHA1 hashes to 20-character hashes which I can then transfer to the new ASP.NET membership data table? I'm pretty familiar with ASP.NET membership by now but I feel like I'm just missing this one piece. However, it may also be known that SHA1 in Ruby and SHA1 in .NET are incompatible, so I'm fighting a losing battle... Thanks in advance for any insight.

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  • Cannot we pass Hash table in WCF services?

    - by Thinking
    I have just odne something like [ServiceContract] public interface IService1 { [OperationContract] Hashtable MyHashTable(); // TODO: Add your service operations here } and implemented as public Hashtable MyHashTable() { Hashtable h = new Hashtable(); for (int i = 0; i < 10; i++) h.Add(i, "val" + i.ToString()); return h; } When I am trying to run the service , I am getting the error "This operation is not supported in WCF test client" Why so? Thanks

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  • are deleted entries counted in the load factor of a hash table using open addressing

    - by Dr. Monkey
    When calculating the load factor of a hashtable with an open-addressing array implementation I am using: numberOfKeysInArray/sizeOfArray however it occurred to me that since deleted entries must be marked as such (to distinguish them from empty spaces), it might make sense to include these in the number of keys. My thinking is that as far as estimating the average number of probes to find an entry, deleted entries should count towards the load factor, but as far as inserting a new key they should not. Which is the proper calculation: including deleted keys or not?

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  • Creating a Rails query from a hash of user input

    - by Jamie
    I'm attempting to create a fairly complex search engine for a project using a variable number of search criteria. The user input is sorted into an array of hashes. The hashes contain the following information: { :column => "", :value => "", :operator => "", # Such as: =, !=, <, >, etc. :and_or => "", # Two possible values: "and" and "or" } How can I loop through this array and use the information in these hashes to make an ActiveRecord WHERE query?

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  • PHP secure logon script - md5 hash is not matching the hash i wrote to the database in a previous sc

    - by Chris Sobolewski
    I am trying to cobble together a login script in PHP as a learning project. This is the code for my database write when the user registers. Both of these values are written to the database. $this->salt = md5(uniqid()); $this->password = md5($password.$salt); Upon logging in, the following function is fired. For some function challengeLogin($submittedPassword, $publicSalt, $storedPassword){ if(md5($submittedPassword.$publicSalt) == $actualPassword){ return 0; }else{ return 1; }; } Unfortunately, on stepping through my code, the two values have never equaled. Can someone help me understand why?

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  • merging arrays of hashes

    - by Ben
    I have two arrays of hashes. Array1 => [{attribute_1 = A, attribute_2 = B}, {attribute_1 = A, attribute_2 = B}] Array2 => [{attribute_3 = C, attribute_2 = D}, {attribute_3 = C, attribute_4 = D}] Each hash in the array is holding attributes for an object. In the above example, there are two objects that I'm working with. There are two attributes in each array for each object How do I merge the two arrays? I am trying to get a single array (there is no way to get a single array from the start because I have to make two different API calls to get these attributes). DesiredArray => [{attribute_1 = A, attribute_2 = B, attribute_3 = C, attribute_2 = D}, {attribute_1 = A, attribute_2 = B, attribute_3 = C, attribute_2 = D}] I've tried a couple things, including the iteration methods and the merge method, but I've been unable to get the array I need.

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  • how to Compute the average probe length for success and failure - Linear probe (Hash Tables)

    - by fang_dejavu
    hi everyone, I'm doing an assignment for my Data Structures class. we were asked to to study linear probing with load factors of .1, .2 , .3, ...., and .9. The formula for testing is: The average probe length using linear probing is roughly Success-- ( 1 + 1/(1-L)**2)/2 or Failure-- (1+1(1-L))/2. we are required to find the theoretical using the formula above which I did(just plug the load factor in the formula), then we have to calculate the empirical (which I not quite sure how to do). here is the rest of the requirements **For each load factor, 10,000 randomly generated positive ints between 1 and 50000 (inclusive) will be inserted into a table of the "right" size, where "right" is strictly based upon the load factor you are testing. Repeats are allowed. Be sure that your formula for randomly generated ints is correct. There is a class called Random in java.util. USE it! After a table of the right (based upon L) size is loaded with 10,000 ints, do 100 searches of newly generated random ints from the range of 1 to 50000. Compute the average probe length for each of the two formulas and indicate the denominators used in each calculationSo, for example, each test for a .5 load would have a table of size approximately 20,000 (adjusted to be prime) and similarly each test for a .9 load would have a table of approximate size 10,000/.9 (again adjusted to be prime). The program should run displaying the various load factors tested, the average probe for each search (the two denominators used to compute the averages will add to 100), and the theoretical answers using the formula above. .** how do I calculate the empirical success?

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  • Ignoring extra keys in a hash passed in to create

    - by denniss
    Does rails provide a way to ignore extra keys that are passed in to create. Supposed User has two attributes, first_name and last_name. When I do User.create({ :first_name => "first", :last_name => "last", :age => 10}) that line gives me an UknonwnAttributeError. Well, that makes sense, it happens cause age is not one of the attributes. But is there a way to just ignore key-value pair that is not one of the attributes for User?

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  • Glib segfault g_free hash table

    - by Mike
    I'm not quite sure why if I try to free the data I get segfault. Any help will be appreciate it. static GHashTable *hashtable; static void add_inv(char *q) { gpointer old_key, old_value; if(!g_hash_table_lookup_extended(hashtable, q, &old_key, &old_value)){ g_hash_table_insert(hashtable, g_strdup(q), GINT_TO_POINTER(10)); }else{ (old_value)++; g_hash_table_insert(hashtable, g_strdup(q), old_value); g_hash_table_remove (hashtable, q); // segfault g_free(old_key); // segfault g_free(old_value); // segfault } } ... int main(int argc, char *argv[]){ hashtable = g_hash_table_new(g_str_hash, g_str_equal); ... g_hash_table_destroy(hashtable); }

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  • accessing a value of a nested hash

    - by st
    Hello! I am new to perl and I have a problem that's very simple but I cannot find the answer when consulting my perl book. When printing the result of Dumper($request); I get the following result: $VAR1 = bless( { '_protocol' => 'HTTP/1.1', '_content' => '', '_uri' => bless( do{\(my $o = 'http://myawesomeserver.org:8081/counter/')}, 'URI::http' ), '_headers' => bless( { 'user-agent' => 'Mozilla/5.0 (X11; U; Linux i686; en; rv:1.9.0.4) Gecko/20080528 Epiphany/2.22 Firefox/3.0', 'connection' => 'keep-alive', 'cache-control' => 'max-age=0', 'keep-alive' => '300', 'accept' => 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8', 'accept-language' => 'en-us,en;q=0.5', 'accept-encoding' => 'gzip,deflate', 'host' => 'localhost:8081', 'accept-charset' => 'ISO-8859-1,utf-8;q=0.7,*;q=0.7' }, 'HTTP::Headers' ), '_method' => 'GET', '_handle' => bless( \*Symbol::GEN0, 'FileHandle' ) }, 'HTTP::Server::Simple::Dispatched::Request' ); How can I access the values of '_method' ('GET') or of 'host' ('localhost:8081'). I know that's an easy question, but perl is somewhat cryptic at the beginning. Thank you, St.

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  • Hash Digest / Array Comparison in C#

    - by Erik Karulf
    Hi All, I'm writing an application that needs to verify HMAC-SHA256 checksums. The code I currently have looks something like this: static bool VerifyIntegrity(string secret, string checksum, string data) { // Verify HMAC-SHA256 Checksum byte[] key = System.Text.Encoding.UTF8.GetBytes(secret); byte[] value = System.Text.Encoding.UTF8.GetBytes(data); byte[] checksum_bytes = System.Text.Encoding.UTF8.GetBytes(checksum); using (var hmac = new HMACSHA256(key)) { byte[] expected_bytes = hmac.ComputeHash(value); return checksum_bytes.SequenceEqual(expected_bytes); } } I know that this is susceptible to timing attacks. Is there a message digest comparison function in the standard library? I realize I could write my own time hardened comparison method, but I have to believe that this is already implemented elsewhere.

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  • Creating a unique key based on file content in python

    - by Cawas
    I got many, many files to be uploaded to the server, and I just want a way to avoid duplicates. Thus, generating a unique and small key value from a big string seemed something that a checksum was intended to do, and hashing seemed like the evolution of that. So I was going to use hash md5 to do this. But then I read somewhere that "MD5 are not meant to be unique keys" and I thought that's really weird. What's the right way of doing this? edit: by the way, I took two sources to get to the following, which is how I'm currently doing it and it's working just fine, with Python 2.5: import hashlib def md5_from_file (fileName, block_size=2**14): md5 = hashlib.md5() f = open(fileName) while True: data = f.read(block_size) if not data: break md5.update(data) f.close() return md5.hexdigest()

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  • Question about the mathematical properties of hashes

    - by levand
    Take a commonly used binary hash function - for example, SHA-256. As the name implies, it outputs a 256 bit value. Let A be the set of all possible 256 bit binary values. A is extremely large, but finite. Let B be the set of all possible binary values. B is infinite. Let C be the set of values obtained by running SHA-256 on every member of B. Obviously this can't be done in practice, but I'm guessing we can still do mathematical analysis of it. My Question: By necessity, C ? A. But does C = A?

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  • Ruby: rules for implicit hashes

    - by flyer
    Why second output shows me only one element of Array? Is it still Array or Hash already? def printArray(arr) arr.each { | j | k, v = j.first printf("%s %s %s \n", k, v, j) } end print "Array 1\n" printArray( [ {kk: { 'k1' => 'v1' }}, {kk: { 'k2' => 'v2' }}, {kk: { 'k3' => 'v3' }}, ]) print "Array 2\n" printArray( [ kk: { 'k1' => 'v1' }, kk: { 'k2' => 'v2' }, kk: { 'k3' => 'v3' }, ]) exit # Output: # # Array 1 # kk {"k1"=>"v1"} {:kk=>{"k1"=>"v1"}} # kk {"k2"=>"v2"} {:kk=>{"k2"=>"v2"}} # kk {"k3"=>"v3"} {:kk=>{"k3"=>"v3"}} # Array 2 # kk {"k3"=>"v3"} {:kk=>{"k3"=>"v3"}}

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