If my computer is a 32 bit system, it has a 32 bit address right? But when I print any memory address in C why do I get an address <32bit?

问题

For example

printf("%u",&a);

gives me output

65524

which is a 16 bit address.

回答1:

Because you used wrong format specifier which invokes undefined behavior.

To print a pointer, you should use %p format specifier and cast the argument to void*. Something like

 printf("%p",(void *)&a);

will do the job.

That said, you should look into the concepts of virtual memory the first thing.

回答2:

You can also simply answer your assumption about address size by checking the size of any pointer instead of a test variable's address:

printf("%zu\n", sizeof(int*));
printf("%zu\n", sizeof(float*));

Assuming that a byte in all systems equals eight bit, you can see about size of address.

Please see this SO post

回答3:

To find the highest possible address for (most of) the currently common systems do this:

#include <stdint.h>
#include <stdio.h>

int main(void)
{
  uintptr_t uip = (uintptr_t) -1;
  void * vp = (void*) uip;

  printf("%p\n", vp);
}

来源：https://stackoverflow.com/questions/37256145/if-my-computer-is-a-32-bit-system-it-has-a-32-bit-address-right-but-when-i-pri