Question

I ran this sample program.

const int x = 5;
int main(int argc, char** argv)
{
    int x[x];

    int y = sizeof(x) / sizeof(int);

    return 0;
}

The Value of y is 5.

But how is this possible? Because when I debugged, the value of x was 5, so the size of x is 4 and size of int is 4. And value of y should have been different .

What am I missing?

Answer 1

sizeof(x) refers to the array which has 5 elements of type int. Hence the output.

The code would be a lot easier to understand if you didn’t overload the name x.

Technically what you have here is a variable length array, a VLA. This is because C const actually means read-only and has to be evaluated at run time. Hence sizeof, in this case, is evaluated at runtime.

If you had used a literal to size your array instead, i.e. int x[5]; then sizeof would have been evaluated at compile time.

If the code had been compiled as C++ then the const would be a true const, and so available for evaluation at compile time.