why answer is 16?

Question

#include <stdio.h>

void main() {
// Write C code here
long double a;
signed char b;
int arr[sizeof(!a+b)];
printf("%d",sizeof(arr));

}
// answer is 16

Answer 1

If you pass !a to sizeof(), everything I explained earlier still holds, since the type of !a is int. However, if you construct the array with

sizeof( a + b )

elements, then the output changes because

a + b

is of type long double, not int, and sizeof( long double ) is 16 on SoloLearn's system. From then on, it applies again that for an array of 16 integers, each occupying 4 byte, the size of the total array would turn out 64.

Answer 2

The logical NOT operator has type int, therefore the expression

!a + b

is an int as well, which typically has a size of 4 byte on 64-bit systems.
As a consequence, "arr" is an array of four integers, making its total size 16 byte, although the exact output depends on the size of int on the system the code is compiled on.

Answer 3

Shadow Thank you so much

Answer 4

But I am a little confused..how is arr consist of 4 intergers. Shadow....in line 3 after main function if I pass !a the ans remains 16 but if I change the parameter to a+b the ans is 64.

Answer 5

Shadow I get it now🤗.thanks