What's the proper use of printf to display pointers padded with 0s

Question

In C, I\'d like to use printf to display pointers, and so that they line up properly, I\'d like to pad them with 0s.

My guess was that the proper way to do this was:

Answer 1

The only portable way to print pointer values is to use the "%p" specifier, which produces implementation-defined output. (Converting to uintptr_t and using PRIxPTR is likely to work, but (a) uintptr_t was introduced in C99, so some compilers may not support it, and (b) it's not guaranteed to exist even in C99 (there might not be an integer type big enough to hold all possible pointer values.

So use "%p" with sprintf() (or snprintf(), if you have it) to print to a string, and then print the string padding with leading blanks.

One problem is that there's no really good way to determine the maximum length of the string produced by "%p".

This is admittedly inconvenient (though of course you can wrap it in a function). If you don't need 100% portability, I've never used a system were casting the pointer to unsigned long and printing the result using "0x%16lx" wouldn't work. [UPDATE: Yes, I have. 64-bit Windows has 64-bit pointers and 32-bit unsigned long.] (Or you can use "0x%*lx" and compute the width, probably something like sizeof (void*) * 2. If you're really paranoid, you can account for systems where CHAR_BIT > 8, so you'll get more than 2 hex digits per byte.)