Question

I want to use variable a from foo.c in main.c, and I write:

foo.c
#include <stdio.h>

int a[] = {3, 2};

void foo()
{
    printf("foo\taddress of a:%x\n", a);
    printf("foo\tvalue of a[0]:%x\n", a[0]);
}

main.c
#include <stdio.h>

extern int *a;

int main(void)
{
    foo();
    printf("main\taddress of a : %x\n", a);
    printf("main\tvalue of a[0] : %x\n", a[0]);

    return 0;
}

and the result output:

foo address of a:804a014
foo value of a[0]:3
main    address of a : 3
Segmentation fault (core dumped)

why?

Answer 1

The type of a is int[2], not int*. Try again with

extern int a[2];

---

The C compiler cannot type-check across source files. Therefore, when you say int* a, the compiler will assume you’re telling the truth, use pointer semantics, and will not issue any compiler error.

There are subtle difference between arrays and pointers. Let’s assume a 32-bit system. Then the content of “a” will be distributed like this:

    a
0x100       0x104      0x108   ← address
    +-----------+----------+
    |         3 |        2 |   ← content
    +-----------+----------+

When a is an array,

• The value of the expression a is will be converted to the address of a. Therefore, when you print a, you will get its address, i.e. “0x100”.
• The operation a[n] in C is equivalent to *(a + n), i.e. advance the address a by n units, and then dereference it to get the content. Therefore, a[0] is equivalent to *0x100, which returns the content at 0x100, i.e. “3”.

When a is a pointer,

• The “value” of a is the content at the provided address. In fact this is the norm, the array type is a special case. Therefore, when you print a, you will get the content at that address, i.e. “3”.
• The operation a[n] is still *(a + n). Therefore, a[0] is equivalent to *3, which causes segmentation fault because the address “3” is invalid.