Why can arrays be assigned directly?

Question

Consider this code snippet:

void foo(int a[], int b[]){
    static_assert(sizeof(a) == sizeof(int*));
    static_assert(sizeof(b) == sizeof(int*));
    b = a;
    printf("%d", b[1]);
    assert(a == b); // This also works!
}

int a[3] = {[1] = 2}, b[1];
foo(a, b);

Output (no compilation error):

2

I can't get the point why b = a is valid. Even though arrays may decay to pointers, shouldn't they decay to const pointers (T * const)?

Answer 1

They can't.

Arrays cannot be assigned to. There are no arrays in the foo function. The syntax int a[] in a function parameter list means to declare that a has type "pointer to int". The behaviour is exactly the same as if the code were void foo(int *a, int *b). (C11 6.7.6.3/7)

It is valid to assign one pointer to another. The result is that both pointers point to the same location.

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Even though arrays may decay to pointers, shouldn't they decay to const pointers (T * const)?

The pointer that results from array "decay" is an rvalue. The const qualifier is only meaningful for lvalues (C11 6.7.3/4). (The term "decay" refers to conversion of the argument, not the adjustment of the parameter).

Answer 2

Quoting C11, chapter §6.7.6.3, Function declarators (including prototypes)

A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. [...]

So, a and b are actually pointers, not arrays.

There's no assignment to any array type happennning here, hence there's no problem with the code.