Array-notation decay to pointer-notation - only for function parameters?

Question

I'm currently learning about how C functions can accept multidimensional arrays, by following the treatment in the book C primer plus, by Stephen Prata (6th edition).

In the book, the author mentions that function that deal with multidimensional array can be declared as...

void somefunction( int (* pt)[4] );

Alternatively, if (and only if) pt is a formal parameter to a function, you can declare it as follows:

void somefunction( int pt[][4] );

I understand that in C, array passed into a function will decay into a pointer to the corresponding type. So pt in int pt[][4] will decay into int (* pt)[4], which is a pointer to array of 4 ints.

But I don't understand why this behavior only happens 'if (and only if) pt is a formal parameter to a function". What does this mean? Why is this the case?

Answer 1

I think he means that you may not declare an array like this

int pt[][4];

in a program (except a declaration with external or internal linkage) because it is an incomplete type that is the size of the array is unknown.

However you may use such a declaration as a parameter declaration

void somefunction( int pt[][4] );

because the parameter is adjusted by the compiler to pointer of the type

void somefunction( int ( *pt )[4] );

and pointers are always complete types.

As result for example these function declarations

void somefunction( int pt[100][4] );
void somefunction( int pt[10][4] );
void somefunction( int pt[][4] );

are equivalent and declare the same one function.

Below there is an example of declaring an array (when a declaration is not a definition) of incomplete type with external linkage.

#include <stdio.h>

int pt[2][4];

int main(void) 
{
    extern int pt[][4];

    printf( "sizeof( pt ) = %zu\n", sizeof( pt ) );

    return 0;
}

The program output is

sizeof( pt ) = 32