在C中打印指针的地址

Question

I'm reproducing printf from scrap and I need to store pointers address into a string then print it, so first I cast void* into an unsigned int then itoa it to hexadecimal but the last three char are wrong.

int     main(void)
{
    char    str[] = "printf from scrap!";

    my_printf("MY_PRINTF:'%p'", (void*)str);
    printf("\n   PRINTF:'%p'\n\n", (void*)str);

    return (0);
}

int     conv_p(va_list args)
{
    void    *ptr;
    unsigned int        ptrint;

    ptr = va_arg(args, void*);
    ptrint = (unsigned int)&ptr;
    my_putstr("0x7fff");
    my_putstr(my_itoa_base_uint(ptrint, 16));

    return (1);
}

Output:

MY_PRINTF:'0x7fff505247b0'
   PRINTF:'0x7fff50524a20'

As you can see the last three char are wrong, is there any documentation about that?

Answer 1

In the second case, you're converting the address of the variable ptr to an int , rather than its value (the pointer you're interested in).

Replacing (unsigned int)&ptr; with (unsigned int)ptr; will give you consistent values.

And an additional aside: there's no guarantee unsigned int is large enough to represent the pointer value: you should use intptr_t or uintptr_t from <stdint.h> .