java help bitwise operations

Question

System.out.println(
  s + ", long: " + l + ", binary: ");
System.out.print("   ");
for(int i = 63; i >= 0; i--)
  if(((1L << i) & l) != 0)
    System.out.print("1");
  else
    System.out.print("0");
System.out.println();

Answer 1

You forgot to update the mask:

    for(int i=1; i<=32; i++) {
        if( (mask & number) != 0 )
            System.out.print(1);
        else
            System.out.print(0);


        if( (i % 4) == 0 )
            System.out.print(" ");

        mask = mask >> 1;
    }

Answer 2

The text of the question has been edited numerous times, so it's hard to tell what the question is, but here are some remarks.

On signed vs unsigned shift

One revision of the question contains this line:

int mask = 1 << 31;

One answer suggests that what was missing is this line:

mask = mask >> 1;

This actually won't work, because >> is a signed shift, and would result in the incorrect mask value for this purpose (since emptied bits are filled with 1s). Any negative number would be converted to 32 bits, all 1.

What is needed is the unsigned right shift.

mask >>>= 1;

Note that compound assignment has been used for brevity. Unsigned right shift >>> fills emptied bits with 0. As used here, it ensures that mask will always have only one bit set, which is what is required for this problem.

See also

• JLS 15.19 Shift Operators
• JLS 15.26.2 Compound Assignment Operators

Similar question

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Alternative solution

There's actually a simpler solution to convert a 32-bit int into 32 bits, separated into groups of 4 bits.

static String zeroes(int length) {
    return (length <= 0) ? ""
      : String.format("%0" + length + "d", 0);
}
//...

int num = 8675309;

// convert to binary
String s = Integer.toBinaryString(num);
System.out.println(s);
// prints "100001000101111111101101"

// fill in leading zeroes
s = zeroes(Integer.SIZE - s.length()) + s;
System.out.println(s);
// prints "00000000100001000101111111101101"

s = s.replaceAll("(?!$)(?<=\\G.{4})", " ");
System.out.println("[" + s + "]");
// prints "[0000 0000 1000 0100 0101 1111 1110 1101]"

Even if this is homework, the variety of techniques used here should still be instructional.

See also

• Integer.toBinaryString
• java.util.Formatter syntax
• regular-expressions.info/Lookarounds and anchors