Question

For example, I need to grab an unknown number, let’s say 3, and find the binary (2^3) – 1 times, from 0 to 111 (0-7). Obviously, the number of digits I need depends on whatever number ‘n’ in 2^n.

So, if the number is 3, I would need the output to be:

000
001
010
011
100
101
111

Now obviously I can do this manually with a String.format(“%03d”, NumberInBinary) operation, but that’s hardcoding it for 3 digits. I need to do the equivalent code with an unknown number of digits, how can I do that? (as in String.format(“%0nd”, yournumber) where n is the number of digits.)

Answer 1

if n = 4, NumberInBinary = 101;

String.format("%0"+n+"d", NumberInBinary);

with output

0101