[ad_1] You’ve been tasked with implementing a hashtable — this involves defining data types and functions to manipulate those types. I have no idea what you’re doing. First define a hashtable — do this by defining two types; one table type and one slot type. typedef struct table table; typedef struct slot slot; The table … Read more
[ad_1] MD5 is a cryptographic hash function. Cryptographic hashes are one way functions. You cannot reverse a cryptographic hash value, but you can brute force messages to find one. Brute forcing means trying all possible input strings and then checking if the hash value is correct. This is possible because cryptographic hashes are also computationally … Read more
[ad_1] I corrected your code also as per your need, and solved further, $ArrayX = [8349310431,8349314513] $ArrayY = [667984788,667987788] $ArrayZ = [148507632380,153294624079] $range_map = $ArrayX.zip([$ArrayY.map(&:to_i), $ArrayZ.map(&:to_i)].transpose).sort $ArrayX = [8349310431,8349314513] => [8349310431, 8349314513] $ArrayY = [667984788,667987788] => [667984788, 667987788] $ArrayZ = [148507632380,153294624079] => [148507632380, 153294624079] $range_map = Hash[$ArrayX.zip([$ArrayY.map(&:to_i), $ArrayZ.map(&:to_i)].transpose).sort] => {8349310431=>[667984788, 148507632380], 8349314513=>[667987788, 153294624079]} keys = … Read more
[ad_1] If you read the documentation, you will learn it’s: key, old value, and new value. h1.merge(h2) { |key, old, new| new – old } means the final result will have value h2[key] – h1[key]. Since you are merging h2 over h1, values from h2 will be new values and those from h1 will be … Read more
[ad_1] $ms = md5($transactionid.’:’.$secretkey); You need to concatenate with . [ad_2] solved Divide md5 function with colon (:) [closed]
[ad_1] Either make the information available by using statics (not recommended), use some kind of database (could be as simple as a text file) or pass an Intent along with your Activity. A nice tutorial on adding information to an Intent is found here: http://startandroid.ru/en/lessons/complete-list/241-lesson-28-extras-passing-data-using-intent.html 0 [ad_2] solved How to access a HashMap from another … Read more
[ad_1] Arrays are allocated as single, large blocks of memory and entries are accessed by their indexes. The order of entries is fixed and they need have no particular identity apart from their position in the array. Other more complex data structures allow one to store objects identified and accessed using some sort of key. … Read more
[ad_1] What is looks like is that you actually have a hash containing some Arrays. Cleaning it up I assume it should look something like this: hash = {:time_frame=>”Today”, :locations=>[“Tampa”, “Atlanta”, “California”, “Georgia”, “South Lake Union”], :local_totals=>[10000.0, 30,000, 70000, 50000, :expenses=>[2000, 10000, 4000, 6000]} Assuming that is correct, you can do something like this to … Read more
[ad_1] You need to join the values in a string: “#{address[‘house_number’]} #{address[‘street_name’]},\n#{address[‘city’]},\n#{address[‘post_code’]}” You could also improve the formatting by making this a helper method, and using a HEREDOC: def formatted_address(address) <<~ADDRESS #{address[‘house_number’]} #{address[‘street_name’]}, #{address[‘city’]}, #{address[‘post_code’]} ADDRESS end Usage: address = { “house_number” => 20, “street_name” => “Mount Park Road”, “city” => “Greenfield”, “post_code” => “WD1 … Read more
[ad_1] The most space efficient way to store binary data is to store it as bytes. The only way you may get it even shorter is via compression. But for 92 Characters that will not amount to much. As for Base64: There are cases where we are forced to transmit binary data over a medium … Read more
[ad_1] With this code you are doing linear probing index = hash(entry.key); while (!is_vacant(index)) index = next_index(index); template <class RecordType> inline std::size_t table<RecordType>::next_index(std::size_t index) const // Library facilities used: cstdlib { return ((index+1) % CAPACITY); } Say your map is nearly full and hash returns 23, then the next slots you are going to test … Read more
[ad_1] I’m not sure what exactly you want to do, but you can try the following: arr = [“Japan”, “Egypt”, “Spain”, “Brazil”] arr.each { |a| Object.const_set(a, Hash.new(a)) } #initialize hashes Country = arr.map {|a| [a, 0]}.to_h #=> {“Japan”=>0, “Egypt”=>0, “Spain”=>0, “Brazil”=>0} or Country = Hash[arr.map {|a| [a, 0]}] #=> {“Japan”=>0, “Egypt”=>0, “Spain”=>0, “Brazil”=>0} Note: Ruby … Read more
[ad_1] array[0].max_by{|i| i[0][‘value’].to_i} Should work 17 [ad_2] solved find maximum in hash ruby [duplicate]
[ad_1] If you can reverse it, then by definition it isn’t a hash. It’s an encoding. Any encoding will have zero collisions (otherwise it wouldn’t be able to accurately reverse it). A common encoding for this purpose is base64. [ad_2] solved What is the best reversable hash algorithm for a URL? (near-Zero collision!) [closed]
[ad_1] A central principle of a hash is that it is one way, as you have read. Since you want to be able to perform a two way conversion between your data and a secured version you need to use encryption, which is specifically designed to both encrypt and then later decrypt the data in … Read more