Question

#include "stdio.h"

#define COUNT(a) (sizeof(a) / sizeof(*(a)))

void test(int b[]) {
  printf("2, count:%d\n", COUNT(b));
}

int main(void) {
  int a[] = { 1,2,3 };

  printf("1, count:%d\n", COUNT(a));
  test(a);

  return 0;
}

The result is obvious:

1, count:3
2, count:1

My questions:

1. Where is the length(count/size) info stored when “a” is declared?
2. Why is the length(count/size) info lost when “a” is passed to the test() function?

Answer 1

There’s no such thing as “array pointer” in C language.

The size is not stored anywhere. a is not a pointer, a is an object of type int[3], which is a fact well known to the compiler at compile time. So, when you ask the compiler to calculate sizeof(a) / sizeof(*a) at compile time the compiler knows that the answer is 3.

When you pass your a to the function you are intentionally asking the compiler to convert array type to pointer type (since you declared the function parameter as a pointer). For pointers your sizeof expression produces a completely different result.