What is an expression of VLA type?

Question

I was trying to understand the working of sizeof operator and I came across this question. Following is the code from that question

#include <stdio.h>

int main() {
    int i = 0;
    int a[i];
    printf("%zu\n",sizeof(a[i++]));
    printf("%d\n",i); // Here, print 0 instead of 1
    return 0;
}

array a here is variable length but when its used as an oprand to the sizeof operator the variable i is not incremented.

some of the comments and answer say that a[i++] is not a VLA type and suggest that op should use a 2D VLA to see the side effect( sizeof evalutes its oprand).

I dont clearly understand why a[i++] doesn't qualify as a VLA expression.i think it has something to do with the fact that we can leave the first dimension unspecified of an array while passing to a fucntion.

so the question is in general what is considered a VLA expression ?

Answer 1

I don't clearly understand why a[i++] doesn't qualify as a VLA expression

a[n] is not a VLA expression, even though a is a VLA, because the expression produces a single element of a VLA, not the VLA itself.

in general what is considered a VLA expression?

In your example, a would be a VLA expression, because it represents a variable-length array.

A 2D VLA would give you a scenario where sizeof would evaluate its operand:

int x, y;
scanf("%d %d", &x, &y);
int a[x][y];
int i = 0;
size_t z = sizeof(a[i++]);
printf("%d %d\n", i, (int)z);

This prints 1, because now not only a is a VLA, but a[n] is a VLA as well (demo).