Question

I understand that in order to return a string from a function I have to return a pointer. What I don’t understand is why a char array is treated somewhat different from, say, integer, and gets destroyed when you exit the function. That’s probably because I come from a high-level languages world, but this seems to be equally valid to me:

int x = 1;
return x;

char x[] = "hello";
return x;

Answer 1

The reason is both simple, yet subtle: C functions cannot be declared to return arrays.

When you return a pointer, like this:

char *foo(void)
{
    char x[] = "hello";
    return x;
}

The return x; is not actually returning x. It is returning a pointer to the first element of x¹ – it is exactly the same as saying:

return &x[0];

It should be more clear why this isn’t correct – it is exactly analagous to this:

int *foo(void)
{
    int x = 100;
    return &x;
}

It is, however, possible to return structures from functions – so you can return an array as long as it wrapped inside a struct. The following is quite OK:

struct xyzzy {
    char x[10];
};

struct xyzzy foo(void)
{
    struct xyzzy x = { "hello" };
    return x;
}

---

^{1. This is a consequence of a special case rule for array types. In an expression, if an array isn’t the subject of either the unary & or sizeof operators, it evaluates to a pointer to its first element. Trying to pin down actual an array in C is a bit like trying to catch fog in your hands – it just disappears, replaced by a pointer.}