Where is `%p` useful with printf?

After all, both these statements do the same thing...

int a = 10;
int *b = &a;
printf("%p\n",b);
printf("%08X\n",b);

For example (with different addresses):

0012FEE0
0012FEE0

It is trivial to format the pointer as desired with %x, so is there some good use of the %p option?

They do not do the same thing. The latter printf statement interprets b as an unsigned int, which is wrong, as b is a pointer.

Pointers and unsigned ints are not always the same size, so these are not interchangeable. When they aren't the same size (an increasingly common case, as 64-bit CPUs and operating systems become more common), %x will only print half of the address. On a Mac (and probably some other systems), that will ruin the address; the output will be wrong.

Always use %p for pointers.

At least on one system that is not very uncommon, they do not print the same:

~/src> uname -m
i686
~/src> gcc -v
Using built-in specs.
Target: i686-pc-linux-gnu
[some output snipped]
gcc version 4.1.2 (Gentoo 4.1.2)
~/src> gcc -o printfptr printfptr.c
~/src> ./printfptr
0xbf8ce99c
bf8ce99c

Notice how the pointer version adds a 0x prefix, for instance. Always use %p since it knows about the size of pointers, and how to best represent them as text.

You cannot depend on %p displaying a 0x prefix. On Visual C++, it does not. Use %#p to be portable.

The size of the pointer may be something different than that of int. Also an implementation could produce better than simple hex value representation of the address when you use %p.