Question

#include <stdio.h>

int main(void) {

int TEST_HB[10]; 
memset(TEST_HB,'9', sizeof( TEST_HB));

printf("%c\n",TEST_HB[9]);

printf ("TEST_HB[10]=%d\n",sizeof( TEST_HB[40]));   // shows 4
printf ("Arraysize=%d\n",(sizeof(int)*10));     // gets the expectected results
return 0;

} 

I believe that sizeof (myArray) should return the total size of the array in bytes. But why does sizeof( TEST_HB[40]) returns 4 when it is not defined ?

Answer 1

TEST_HB[40] is an expression with type int (and undefined behavior if evaluated, since 40 is too big for the array). So sizeof(TEST_HB[40]) is the size of an int on your implementation: 4 is typical.

Importantly, sizeof does not evaluate its operand unless it’s a VLA — it just uses the type. Therefore your code has defined behavior even though there’s no such object as TEST_HB[40].

Actually, I say it has defined behavior, but sizeof evaluates to type size_t, which isn’t printed with %d. Use %zu where available or consult your compiler documentation. You’ve got away with it here because you’ve got lucky with the varargs calling convention, plus either size_t is the same size as int in your implementation, or your implementation is little-endian, or both.

A fairly common use of the fact that sizeof does not evaluate the operand is to write something like this:

struct Foo *foo = malloc(sizeof(*foo) * number_of_foos_required);

sizeof(*foo) is the same as sizeof(struct Foo), but to some people it’s more “obviously” the right size to use. foo is uninitialized before the memory is allocated, so it’s just as well that sizeof doesn’t actually use the value of foo.

A less common use, that demonstrates the behavior of sizeof:

int i = 0;
printf("%d\n", i);
printf("%d\n", (int)(sizeof(i++))); // i++ is not executed
printf("%d\n", i);