为什么-2147483648在适合int时会自动提升为long？

Question

#include <stdio.h>

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

The above code gives as output (gcc):

8  
4  

Why is -2147483648 automatically promoted to long in 1 ^st printf even when it can fit in an int ?

Also, I tried the same in MinGW and it gives the output:

4  
4  

Can someone please explain what's going on?

Answer 1

The number 2147483648 is too large to fit into an int , so it is promoted to long .

Then, after the number has already been promoted to long , its negative is computed, yielding -2147483648.

If you're curious, you can look at limits.h . On my platform, which uses glibc,

#  define INT_MIN       (-INT_MAX - 1)
#  define INT_MAX       2147483647

On MinGW, sizeof(long) == 4 , so promotion to long won't cut it. According to the C11 standard, the value must be promoted to long long . This doesn't happen, probably because your MinGW compiler is defaulting to C90 or earlier.

Answer 2

-2147483648 is the integral constant expression 2147483648 with the unary minus operator applied.

Your systems both appear to have int and long as 32-bit, and long long as 64-bit.

Since the base-10 constant 2147483648 cannot fit in long int , in C99 it has type long long int . However, C90 did not have this type, so in C90 it has type unsigned long int .

A common workaround for this issue is, as you say, to write -2147483647 - 1 . You will probably see something similar as the definition of INT_MIN if you check your system's limits.h .

Regarding the output 4 , this suggests that on your MinGW system you are using C90 and on your other system you are using C99 or C11 (or C++, or GCC extensions).