printf() formatting for hexadecimal

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\n", i);  // gives 0x00000007printf("0x%08x\n", i);  // gives 0x00000007printf("%#08x\n", i);   // gives 0x000007

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

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

The "0x" counts towards the eight character count. You need "%#010x".

Note that # does not append the 0x to 0 - the result will be 0000000000 - so you probably actually should just use "0x%08x" anyway.

The %#08X conversion must precede the value with 0X; that is required by the standard. There's no evidence in the standard that the # should alter the behaviour of the 08 part of the specification except that the 0X prefix is counted as part of the length (so you might want/need to use %#010X. If, like me, you like your hex presented as 0x1234CDEF, then you have to use 0x%08X to achieve the desired result. You could use %#.8X and that should also insert the leading zeroes.

Try variations on the following code:

#include <stdio.h>int main(void){    int j = 0;    printf("0x%.8X = %#08X = %#.8X = %#010x\n", j, j, j, j);    for (int i = 0; i < 8; i++)    {        j = (j << 4) | (i + 6);        printf("0x%.8X = %#08X = %#.8X = %#010x\n", j, j, j, j);    }    return(0);}

On an RHEL 5 machine, and also on Mac OS X v10.7.5 (Lion), the output was:

0x00000000 = 00000000 = 00000000 = 00000000000x00000006 = 0X000006 = 0X00000006 = 0x000000060x00000067 = 0X000067 = 0X00000067 = 0x000000670x00000678 = 0X000678 = 0X00000678 = 0x000006780x00006789 = 0X006789 = 0X00006789 = 0x000067890x0006789A = 0X06789A = 0X0006789A = 0x0006789a0x006789AB = 0X6789AB = 0X006789AB = 0x006789ab0x06789ABC = 0X6789ABC = 0X06789ABC = 0x06789abc0x6789ABCD = 0X6789ABCD = 0X6789ABCD = 0x6789abcd

I'm a little surprised at the treatment of 0; I'm not clear why the 0X prefix is omitted, but with two separate systems doing it, it must be standard. It confirms my prejudices against the # option.

The treatment of zero is according to the standard.

ISO/IEC 9899:2011 §7.21.6.1 The fprintf function

¶6 The flag characters and their meanings are:...# The result is converted to an "alternative form". ... For x (or X)conversion, a nonzero result has 0x (or 0X) prefixed to it. ...

(Emphasis added.)

Note that using %#X will use upper-case letters for the hex digits and 0X as the prefix; using %#x will use lower-case letters for the hex digits and 0x as the prefix. If you prefer 0x as the prefix and upper-case letters, you have to code the 0x separately: 0x%X. Other format modifiers can be added as needed, of course.

For printing addresses, use the <inttypes.h> header and the uintptr_t type and the PRIXPTR format macro:

#include <inttypes.h>#include <stdio.h>int main(void){    void *address = &address;  // &address has type void ** but it converts to void *    printf("Address 0x%.12" PRIXPTR "\n", (uintptr_t)address);    return 0;}

Example output:

Address 0x7FFEE5B29428

Choose your poison on the length — I find that a precision of 12 works well for addresses on a Mac running macOS. Combined with the . to specify the minimum precision (digits), it formats addresses reliably. If you set the precision to 16, the extra 4 digits are always 0 in my experience on the Mac, but there's certainly a case to be made for using 16 instead of 12 in portable 64-bit code (but you'd use 8 for 32-bit code).