Why is the sizeof operator not evaluated in a for loop condition?

user_9346366 0
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I don't know why the sizeof operator is not evaluated in a for loop condition at run time. I am trying this simple code with different C compilers but it always print nothing. But if I replace sizeof(i) with 4 then it works fine:

for(int i = -2; i <= 4; i++)

#include <stdio.h>

int main()
{

    for(int i = -2; i <= sizeof(i); i++)
        printf("Hello World");

    return 0;
}
sizeof c
4 Answers

The problem is , the result of sizeof() operator is of type size_t, which is an unsigned type.

Next, in the comparison, i <= sizeof(i) as per the usual arithmetic conversion rules, -2, which is a signed value, gets promoted to an unsigned value, producing a huge value, evaluating the condition to false. So the loop condition is not satisfied and the loop body is not executed.

Run your program through a debugger and see the values in each step, it'll be more clear to you once you see the promoted values in the comparison.

user_21739170

sizeof yields a value of unsigned type variety (size_t). The i is converted to that type and the comparison executed as

(size_t)-2 <= 4

something like 4000000000 < 4

user_253240

you need to typecast sizeof(i) into int. that should solve the problem.

so just replace for(int i = -2; i <= sizeof(i); i++) with for(int i = -2; i <= (int) sizeof(i); i++)

user_81929140

sizeof() returns size_t which is an unsigned type. You can try this code instead:

#include<stdio.h>
int main()
{
  int i=-2,j;
  j=(int)sizeof(i);
  for(i=-2;i<=j;i++)
     printf("HELLO\n");
  return 0;
}

You can typecast sizeof() to int.

user_137350940
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