Why does C disallow an array return type?

Question

Why is this valid in C

int * foo(int a,int b){
...
}

but this is invalid

int [] foo(int a,int b){
...
}

Answer 1

The syntax is somewhat funny.

int foo(int a, int b) []
{
    ...
}

But it's not allowed anyway. See n1256 6.7.5.3 paragraph 1, "Function declarators".

A function declarator shall not specify a return type that is a function type or an array type.

You can return pointers to arrays:

int (*foo(int a, int b)) []; // Weird syntax, n'est-ce pas?

But you might as well just return a pointer instead, since the following are equivalent:

int array[] = { ... };
int *x = array, (*y)[] = &array;
x[0];
(*y)[0]; // same as above, just with more cumbersome syntax

Typically, if a function needs to return an array of int, you either return the pointer or pass the pointer in. One of the following:

int *func(int a, int b); // Allocated by func
void func(int a, int b, int *array); // Allocated by caller
void func(int a, int b, int **array); // Allocated by func

The "struct-hack" also works for arrays that have a fixed size:

struct { int arr[50]; } array_struct;
struct array_struct func(int a, int b);

But this is not recommended unless the array is small.

Rationale:

Arrays are often large, and often have a size not known until runtime. Since parameters and return values are passed using the stack and registers (on all ABIs I know of), and the stack has a fixed size, it is somewhat dangerous to pass such a large object around on the stack. Some ABIs also don't gracefully handle large return values, potentially causing extra copies of return values to be generated.

The following code can also be dangerous:

void func(...)
{
    int arr[BIG_NUMBER]; // potential for stack overflow
    int *ptr = alloca(sizeof(int) * BIG_NUMBER); // potential for stack overflow
}