c - printf() formatting for hexadecimal

Question

Why, when printing a number in hexadecimal as an 8 digit number with leading zeros, does %#08X not display the same result as 0x%08X?

When I try to use the former, the 08 formatting flag is removed, and it doesn't work with just 8.

Answer 1

The # part gives you a 0x in the output string. The 0 and the x count against your "8" characters listed in the 08 part. You need to ask for 10 characters if you want it to be the same.

int i = 7;

printf("%#010x
", i);  // gives 0x00000007
printf("0x%08x
", i);  // gives 0x00000007
printf("%#08x
", i);   // gives 0x000007

Also changing the case of x, affects the casing of the outputted characters.

printf("%04x", 4779); // gives 12ab
printf("%04X", 4779); // gives 12AB