sizeof(2147483648) is 8 bytes while sizeof(2147483647+1) is 4 bytes

Question

#include<stdio.h>

int main()
{
    printf("%d\n", sizeof(2147483648));
    printf("%d"  , sizeof(2147483647+1)); 
    return 0;
}

Output:

8  
4

I understand that sizeof(2147483648) is 8 bytes as it cannot fit in 4 bytes and is promoted to long long int. But I do not understand what happens in case of sizeof(2147483647+1)

I found a similar question but it does not discuss the second case.

Answer 1

The rules of integer constant in C is that an decimal integer constant has the first type in which it can be represented to in: int, long, long long.

2147483648

does not fit into an int into your system (as the maximum int in your system is 2147483647) so its type is a long (or a long long depending on your system). So you are computing sizeof (long) (or sizeof (long long) depending on your system).

2147483647

is an int in your system and if you add 1 to an int it is still an int. So you are computing sizeof (int).

Note that sizeof(2147483647+1) invokes undefined behavior in your system as INT_MAX + 1 overflows and signed integer overflows is undefined behavior in C.

Note that while generally 2147483647+1 invokes undefined behavior in your system (INT_MAX + 1 overflows and signed integer overflows is undefined behavior in C), sizeof(2147483647+1) does not invoke undefined behavior as the operand of sizeof in this case is not evaluated.