Convert number to binary string with full padding?

Question

I have a long variable in java and convert it to a binary string like

long var = 24; Long.toBinaryString (shaft);

Now it prints only 7 bits, but I need to display all 64 bits, i.e. all leading zeros, how can I achieve this?

The reason is because I need to iterate over each bit and perform the operation according to the status, is there a better way to do this?

Answer 1

If you want to iterate over bits, you might be better off testing each bit without converting it to a string:

if ((val & (1L << bitPosition)) != 0)
    // etc

, , , :

string padding = "0000000000000000000000000000000000000000000000000000000000000000";
string result = padding + Long.toBinaryString(val);
result = result.substring(result.length() - 64, result.length());  // take the right-most 64 digits

Answer 2

long. "i" "" "".

long i = 1024;
for(int n = 63; n >= 0; n--)
    System.out.println(n + ": " + ((i & (1L << n)) != 0));

Answer 3

, , :

int i=0, thisbit;
mask = 1;
while (i++ < 64)
{
    thisbit = var & mask;
    // check thisbit here...
    //
    var = var >>> 1;
    mask*=2;
}

Answer 4

@Robert. Long.numberOfLeadingZeros(long).

.

Answer 5

, . http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html .

Otherwise use Formatter: http://java.sun.com/j2se/1.5.0/docs/api/java/util/Formatter.html

Answer 6

This will work;

String s =  Long.toBinaryString(val);
while (s.length() < 64)
{
    s = "0" + s;
}

If you want to do this with a StringBuffer, then:

StringBuffer s =  new StringBuffer(Long.toBinaryString(val));
while (s.length() < 64)
{
    s.insert(0, "0");
}
String str = s.toString();