After many calculations I have String containing binary representation of some data. How to convert it into array of ints in a consistent way.
I can take every 32 bits and put it into array but it doesn't seem like this solution.
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3 Answers
Parsing each 32 bit section seems like a perfectly good solution to me. Note that it's probably best to use Long to do the parsing, to avoid problems if the leading digit is 1. For example:
public static int[] parseBinaryToIntArray(String input) {
// TODO: Validation
int[] output = new int[input.length() / 32];
for (int i = 0; i < output.length; i++) {
String section = input.substring(i * 32, (i + 1) * 32);
output[i] = (int) Long.parseLong(section, 2);
}
return output;
}
6 Comments
Jon Skeet
@RobertKilar: What exception and on what input? Hint: the more information you provide in the question, the more likely you are to get answer that work first time...
Jon Skeet
@RobertKilar: Fixed - had become confused between substring in .NET (second argument is the count) and Java (second argument is the end index, exclusive). My point about being more helpful in your question and comments stands though. Please read tinyurl.com/so-hints
Yoda
Mister Jon Skeet. Do you have in possesion the DSA Java implementation without libraries? I have made RSA, SHA-1(I needed this converting binary into int[] for SHA-1) implementation(I can share) last days and I just can't struggle anymore with that DSA- Digital Signature Algorithm. I know what is it for etc. but as I said it's too much. Please help me.
Jon Skeet
@RobertKilar: No - but I'd just use the standard implementation within the Java libraries. Why would you do anything else?
Yoda
I need to practice it before exam. I've already used MD5,SHA-1,RSA,DSA from libraries, but I have to do it without it. I have already done the SHA-1 the RSA and it took 4 days(I looked for sth in the web, but finally I got to do it myself). I am just simply exhausted so I need some implementation from the web.
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Corrected:
public static int[] convertIntoIntArray(String message) {
System.out.println(message.length());
int[] t = new int[message.length()/32];
for(int i = 0; i < t.length; i++){
//t[i] = Integer.parseInt(message.substring(i*32, (i+1)*32),2);
String section = message.substring(i*32, (i+1)*32);
t[i] = (int) Long.parseLong(section, 2);
// System.out.println(i+"fsa");
}
return t;
}
1 Comment
Jon Skeet
Any reason for posting this after my earlier corrected version (which doesn't have all the commented out bits etc)?
Simple example:
String binary = "1001";
int base = 2;
int decimal = Integer.parseInt(binary, base);
System.out.println("Decimal is" + decimal);
This prints:
Decimal is 9
