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):

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

)?

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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.

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).

Why can arrays be assigned directly?

They can't.

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