Question

When you create an integer with leading zeros, how does C handle it? Is it different for different versions of C?

In my case, they just seem to be dropped (but maybe that is what printf does?):

#include <stdio.h>

int main() {
    int a = 005;
    printf("%i\n", a);
    return 0;
}

I know I can use printf to pad with 0s, but I am just wondering how this works.

Answer 1

Leading zeros indicate that the number is expressed in octal, or base 8; thus, 010 = 8. Adding additional leading zeros has no effect; just as you would expect in math, x + 0*8^n = x; there’s no change to the value by making its representation longer.

One place you often see this is in UNIX file modes; 0755 actually means 7*8^2+5*8+5 = 493; or with umasks such as 0022 = 2*8+2 = 10.

atoi(nptr) is defined as equivalent to strtol(nptr, (char **) NULL, 10), except that it does not detect errors – as such, atoi() always uses decimal (and thus ignores leading zeros). strtol(nptr, anything, 0) does the following:

The string may begin with an arbitrary
amount of white space (as determined
by isspace(3)) followed by a single
optional '+' or '-' sign. If base is
zero or 16, the string may then
include a "0x" prefix, and the number
will be read in base 16; otherwise, a
zero base is taken as 10 (decimal)
unless the next character is '0', in
which case it is taken as 8 (octal).

So it uses the same rules as the C compiler.