c++ - General rules of passing/returning reference of array (not pointer) to/from a function?

Question

We can pass reference of an array to a function like:

void f(int (&a)[5]);

int x[5];
f(x);     //okay
int y[6];
f(y);     //error - type of y is not `int (&)[5]`.

Or even better, we can write a function template:

template<size_t N>
void f(int (&a)[N]); //N is size of the array!

int x[5];
f(x);     //okay - N becomes 5
int y[6];
f(y);     //okay - N becomes 6

---

Now my question is, how to return reference of an array from a function?

I want to return array of folllowing types from a function:

int a[N];
int a[M][N];
int (*a)[N];
int (*a)[M][N];

where M and N is known at compile time!

What are general rules for passing and returning compile-time reference of an array to and from a function? How can we pass reference of an array of type int (*a)[M][N] to a function?

EDIT:

Adam commented : int (*a)[N] is not an array, it's a pointer to an array.

Yes. But one dimension is known at compile time! How can we pass this information which is known at compile time, to a function?

Answer 1

If you want to return a reference to an array from a function, the declaration would look like this:

// an array
int global[10];

// function returning a reference to an array
int (&f())[10] {
   return global;
}

The declaration of a function returning a reference to an array looks the same as the declaration of a variable that is a reference to an array - only that the function name is followed by (), which may contain parameter declarations:

int (&variable)[1][2];
int (&functionA())[1][2];
int (&functionB(int param))[1][2];

Such declarations can be made much clearer by using a typedef:

typedef int array_t[10];

array_t& f() {
   return global;
}

If you want it to get really confusing, you can declare a function that takes a reference to an array and also returns such a reference:

template<int N, int M>
int (&f(int (&param)[M][N]))[M][N] {
   return param;
}

Pointers to arrays work the same, only that they use * instead of &.