Length of array in function argument

Question

This is well known code to compute array length in C:

sizeof(array)/sizeof(type)

But I can\'t seem to find out the length of the array pass

Answer 1

sizeof only works to find the length of the array if you apply it to the original array.

int a[5]; //real array. NOT a pointer
sizeof(a); // :)

However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.

int a[5];
int * p = a;
sizeof(p); // :(

As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)

There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?

A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)

A second idea is storing the array length next to the array, just like any modern programming language does:

a -> [5];[0,0,0,0,0]

But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:

struct {
    size_t length;
    int * elements;
}

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Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?