CODEWITHSUNDEEP
javabinary
doublemc
doublemc

Reputation: 3301

Converting int/long to 64-bit binary

I'm looking for a better way to convert an int to 64 bit binary represented as String. Right now I'm converting int to 32bit binary and then adding 32 zeroes in front of it. I'm implementing SHA-1 and I need that int to be 64bit binary.

private static String convertIntTo64BitBinary(int input) {
    StringBuilder convertedInt = new StringBuilder();
    for(int i = 31; i >= 0; i--) { // because int is 4 bytes thus 32 bits
        int mask = 1 << i;
        convertedInt.append((input & mask) != 0 ? "1" : "0");
    }
    for(int i = 0; i < 32; i++){
        convertedInt.insert(0, "0");
    }
    return convertedInt.toString();
}

EDIT 1: Unfortunately this doesn't work:

    int messageLength = message.length();
    String messageLengthInBinary = Long.toBinaryString((long) messageLength);

messageLengthInBinary is "110"

EDIT 2: To clarify: I need the string to be 64 chars long so need all the leading zeros.

Upvotes: 1

Views: 3204

Answers (1)

Justinw
Justinw

Reputation: 661

You could use the long datatype which is a 64 bit signed integer. Then calling Long.toBinaryString(i) would give you the binary representation of i

If you already have an int then you could cast it to a long and use the above.

This doesn't include the leading 0s so the resulting string needs padding to the 64 characters. Something like this would work.

String value = Long.toBinaryString(i)
String zeros = "0000000000000000000000000000000000000000000000000000000000000000" //String of 64 zeros
zeros.substring(value.length()) + value;

See How can I pad a String in Java? for more options.

Upvotes: 3

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