Howto print Long in Binary?

Question

I am trying to print a Long in Binary, but it keeps cutting off 0's. Is there a way to force it to show all bits?

This is my code:

long l = 1;
System.out.println(Long.toBinaryString((long)l));

Returns as mentioned only 1 due to removed 0's i wish to maintain:

0000 0000 0000 0000 0000 0000 0000 0001

Thanks in advance.

My temporary nasty solution:

public String fillZeros(Long value) 
{
    String str = Long.toBinaryString((long) value);
    String temp;
    temp = str;
    while(temp.length() < 32) {
        temp = "0" + temp;
    }
    return temp;
}

possible duplicate of [How to convert a long to a fixed-length 16-bit binary string?](http://stackoverflow.com/questions/2621017/how-to-convert-a-long-to-a-fixed-length-16-bit-binary-string) — dogbane, May 01 '12 at 17:05
At least use a `StringBuilder` because concatenating "0" repeatedly is inefficient. And you're supposed to append up to 64 zeroes, not 32. — Tudor, May 01 '12 at 17:23
@Tudor, thats not entirely true. It depends on if were speaking of `long` in terms of Java or C. Im working with C datatypes, as im only testing on Java. In terms of C `long` is 32-bit and `long long` is 64-bit. — JavaCake, May 01 '12 at 17:28

twain249 · Accepted Answer · 2012-05-01T17:25:13.813

8

you can do this

for(int i = 0; i < Long.numberOfLeadingZeros((long)l); i++) {
      System.out.print('0');
}
System.out.println(Long.toBinaryString((long)l));

That will get what you want except without the spaces between every 4 numbers (you should be able to code that though). There might be a way to do this automatically with a Formatter but I couldn't find it.

Edit:

you can use String's format method if you know the number of 0's you need (I forgot to change it back into a number this will fix the Exception).

String.format("%032d", new BigInteger(Long.toBinaryString((long)l)));

edited May 01 '12 at 17:25

answered May 01 '12 at 17:11

twain249

5,528
1
17
25

Exception in thread "main" java.util.IllegalFormatConversionException: d != java.lang.String – JavaCake May 01 '12 at 17:19
@JavaCake Fixed the exception – twain249 May 01 '12 at 17:25
Also I used 32 because that's how many bits you had in your question but a Long in java is 64 bits so it really should be `"%064d"` – twain249 May 01 '12 at 17:27
That did the trick. I mentioned why im using 32-bit in a comment on the top. – JavaCake May 01 '12 at 17:33

score 0 · Answer 2 · answered May 01 '12 at 17:27

0

If you're willing to pull in Apache Commons StringUtil, you can do this:

    long l = 1;
    System.out.println( StringUtils.leftPad( Long.toBinaryString( l ), 64, "0" ) );

answered May 01 '12 at 17:27

Chris Kessel

5,053
3
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1

Thanks for your answer, but unfortunately im not using StringUtil. – JavaCake May 01 '12 at 17:35
StringUtil has a bunch of other useful stuff, I'd suggest considering adding it to your toolset in general. Unless this is a homework assignment sort of item. – Chris Kessel May 01 '12 at 18:06

score -1 · Answer 3 · answered Mar 23 '18 at 17:42

-1

long value=4;
String mask = "00000000000000000000000000000000";
String str = Long.toBinaryString((long) value);
System.out.println(mask.substring(0, mask.length()-str.length())+str);

For a value of 4 the result will be 00000000000000000000000000000100

answered Mar 23 '18 at 17:42

RustyTheCat

1

Simplest way I know. – RustyTheCat Mar 23 '18 at 17:44

Howto print Long in Binary?

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