Why does a C array decay to a pointer when passed with size

Question

I understand why an array decays to a pointer when passed to a function without specifying its size, eg.:

void test(int array[]);

Why does it do so when passed with the size? eg.

void test(int array[3]);

I am having trouble with sizeof under the latter function signature, which is frustrating as the array length is clearly known at compile time.

Answer 1

void test(int* array);
void test(int array[]);
void test(int array[3]);

All these variants are the same. C just lets you use alternative spellings but even the last variant explicitly annotated with an array size decays to a pointer to the first element.

That is, even with the last implementation you could call the function with an array of any size:

void test(char str[10]) { }

test("test"); // Works.
test("let's try something longer"); // Still works.

There is no magic solution, the most readable way to handle the problem is to either make a struct with the array + the size or simply pass the size as an additional parameter to the function.

LE: Please note that this conversion only applies to the first dimension of an array. When passed to a function, an int[3][3] gets converted to an int (*)[3] , not int ** .

Answer 2

Because you would not be able to call such a function properly. In almost all contexts an array decays to a pointer to the first element, and at any attempt to call the function an array would first be converted to a pointer and the call would be a mismatch. The only contexts where an array does not decay are the operators sizeof , _Alignof , _Alignas and & . These are easier to detect syntactically.