C pointer notation compared to array notation: When passing to function

Solution 1:

When you declare a function parameter as an array, the compiler automatically ignores the array size (if any) and converts it to a pointer. That is, this declaration:

int foo(char p[123]);

is 100% equivalent to:

int foo(char *p);

In fact, this isn't about notation but about the actual type:

typedef char array_t[42];
int foo(array_t p);  // still the same function

This has nothing to do with how you access p within the function. Furthermore, the [] operator is not "array notation". [] is a pointer operator:

a[b]

is 100% equivalent to:

*(a + b)

Solution 2:

There is no real functional difference between the two notations. In C, when you pass an array variable to a function, it decays to a pointer regardless of the notation. However, in my opinion, the pointer notation is preferable. The problem with [] notation in function definitions is that, in my opinion, it is somewhat misleading:

void foo(int array[])
{

}

A ubiquitous mistake among novice C programmers is to assume that sizeof(array) will give you the number of elements in the array multiplied by sizeof(int), like it would if array were an array variable declared on the stack. But the reality is that array has been decayed to a pointer, despite the misleading [] notation, and so sizeof(array) is going to be sizeof(int*). array is really just a pointer to the first element, or possibly a pointer to a single integer allocated anywhere.

For example, we could call foo like this:

int x = 10;
foo(&x);

In which case the [] notation in the definition of foo is kind of misleading.