Why are the addresses of argc and argv 12 bytes apart?

Question

I ran the following program on my computer (64-bit Intel running Linux).

#include <stdio.h>  void test(int argc, char **argv) {     printf("[test] Argc Pointer: %p\n", &argc);     printf("[test] Argv Pointer: %p\n", &argv); }  int main(int argc, char **argv) {     printf("Argc Pointer: %p\n", &argc);     printf("Argv Pointer: %p\n", &argv);     printf("Size of &argc: %lu\n", sizeof (&argc));     printf("Size of &argv: %lu\n", sizeof (&argv));     test(argc, argv);     return 0; } 

The output of the program was

$ gcc size.c -o size $ ./size Argc Pointer: 0x7fffd7000e4c Argv Pointer: 0x7fffd7000e40 Size of &argc: 8 Size of &argv: 8 [test] Argc Pointer: 0x7fffd7000e2c [test] Argv Pointer: 0x7fffd7000e20 

The size of the pointer &argv is 8 bytes. I expected the address of argc to be address of (argv) + sizeof (argv) = 0x7ffed1a4c9f0 + 0x8 = 0x7ffed1a4c9f8 but there is a 4 byte padding in between them. Why is this the case?

My guess is that it could be due to memory alignment, but I am not sure.

I notice the same behaviour with the functions I call as well.

Answer 1

On your system, the first few integer or pointer arguments are passed in registers and have no addresses. When you take their addresses with &argc or &argv, the compiler has to fabricate addresses by writing the register contents to stack locations and giving you the addresses of those stack locations. In doing so, the compiler chooses, in a sense, whatever stack locations happen to be convenient for it.