Does "int (*)[]" decay into "int **" in a function parameter?

Question

I posted this question on programmers.stackexchange earlier today. I have always assumed that int (*)[] does not decay into int ** in function parameters but I got multiple responses to my question that suggested that it does.

I have used int (*)[] heavily in my function parameters but now I have become really confused.

When I compile this function using gcc -std=c99 -pedantic -Wall

void function(int (*a)[])
{
    sizeof(*a);
}

I get this error message:

c99 -Wall -pedantic -c main.c -o main.o
main.c: In function ‘function’:
main.c:3:11: error: invalid application of ‘sizeof’ to incomplete type ‘int[]’ 
make: *** [main.o] Error 1

Which suggests that *a has the type int [] and not int *.

Can someone explain if things like int (*)[] decays into int ** in function parameters and give me some reference (from the standard documents perhaps) that proves why it is so.

Answer 1

Only array types converted to pointer to its first element when passed to a function. a is of type pointer to an array of int, i.e, it is of pointer type and therefore no conversion.

For the prototype

void foo(int a[][10]);

compiler interpret it as

void foo(int (*a)[10]);  

that's because a[] is of array type. int a[][10] will never be converted to int **a. That said, the second para in that answer is wrong and misleading.

As a function parameter, int *a[] is equivalent to int ** this is because a is of array type .

Answer 2

int (*)[] is a pointer to an array of int.

In your example, *a can decay to int*. But sizeof(*a) doesn't do decaying; it is essentially sizeof(int[]) which is not valid.

a can not decay at all (it's a pointer).