CODEWITHSUNDEEP
java
ayush
ayush

Reputation:

How to convert a long to a fixed-length 16-bit binary string?

Hi i want to convert a long integer to binary but the problem is i want a fixed 16 bit binary result after conversion like if i convert 2 to 16 bit binary it should give me 0000000000000010 as ans can anyone help me ?

Upvotes: 6

Views: 26857

Answers (8)

Shashank Shekhar
Shashank Shekhar

Reputation: 4198

I had to do it for a 32 bit number and ended up with:

String stringWord = Long.toBinaryString(word);
while (stringWord.length() < 32) // ensure that length of word is 32
        stringWord = "0" + stringWord;

Upvotes: 2

dckrooney
dckrooney

Reputation: 3121

In terms of an algorithm to convert base10 numbers to binary, I personally think the following is pretty straightforward:

char[] array;

for (i; i < 16; i++)
{
    if (yourNumber % 2 == 0)
          array[16-i] = '0';
    else if (yourNumber % 2 == 1)
          array[16-i] = '1';
    yourNumber = yourNumber / 2;
}

You can then convert your char array to a String if you like.

Upvotes: 1

Thirler
Thirler

Reputation: 20760

In contrast to many suggestions here: Integer.toBinaryString, doesn't work for a 16 bit (a short) and it will not print leading zero's. The reason is that (as the name suggests) this will only work for integers. And for negative numbers the bit representation will change (the first bit indicates a negative number). The two numbers below represent the same number in short and int. So if you want to represent the raw bits you have received (this is the general application of your problem), this function will generate strange output.

decimal: -3
short:                     1111 1111 1111 1101
int:   1111 1111 1111 1111 1111 1111 1111 1101

EDIT: Changed the number above

Hence you can not cast the short if you are interested in the bit.

Java doesn't provide the implementation for short, so you will have to provide your own. Something like this (size is the number of bits):

int displayMask = 1 << (size - 1);
StringBuffer buf = new StringBuffer( size);
for ( int c = 1; c <= size; c++ ) 
{
    buf.append( ( value & displayMask ) == 0 ? '0' : '1' );
    value <<= 1;
}

Upvotes: 3

PATRY Guillaume
PATRY Guillaume

Reputation: 4337

if you want the binary representation of a long, then there is a method in the Long objet to do so :

String Long.toString(long i, int radix);

with a radix of 2, you should have a binary representation.

regards
Guillaume

Upvotes: 0

malaverdiere
malaverdiere

Reputation: 1537

I'm presuming that you want a String output of fixed length (16). Here's what the code would look like: 

String binarized = Integer.toBinaryString(i);
int len = binarized.length();
String sixteenZeroes = "00000000000000000";
if (len < 16)
  binarized = sixteenZeroes.subString(0, 16-len).concat(binarized);
else
  binarized = binarized.subString(len - 16);
return binarized;

Warning: I didn't compile or run it, so make sure no bug is there :)

Upvotes: 4

polygenelubricants
polygenelubricants

Reputation: 384016

Integer.toBinaryString will convert an int to its binary representation as a string.

It does not give you leading zeroes, so if you really need the string to have those and be 16 bits, you can just add them yourself.

You should know how to do that.

Do note that an int is actually 32 bits in Java. You should also know how two's complement works. The binary representation of -1, for example, is 32 1s.

Upvotes: 0

uckelman
uckelman

Reputation: 26264

Most likely what you want is Integer.toBinaryString(), combined with something to ensure that you get exactly 16 places:

int val = 2;
String bin = Integer.toBinaryString(0x10000 | val).substring(1);

The idea here is to get the zero padding by putting a 1 in the 17th place of your value, and then use String.substring() to chop off the leading 1 this creates, thus always giving you exactly 16 binary digits. (This works, of course, only when you are certain that the input is a 16-bit number.)

Upvotes: 23

kgiannakakis
kgiannakakis

Reputation: 104196

Binary is a representation and not a format to convert an integer to. For example, if you have an integer:

int i = 2;

The binary representation will be 00000010. Java has only signed integers, so this link will be helpful.

Upvotes: -1

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