Compare overflow(underflow) number

Question

I am trying to understand the following code:

#include <stdio.h>

int main()
{
    int a=2147483647,b=1;
    printf("%d\n",a+b);
    printf("%d\n",a+b>a);

    return 0;
}

Since a is the maximum integer, so a+b should be negative. But the last line print out "True".

If I change the second print out to printf("%d\n",2147483647+1>2147483647); Their will be an warning and print out false which is what I expected. Is their any setting in the compiler leads to this result?

Answer 1

Integer overflow on signed types is Undefined Behaviour (UB) and the compiler is free to do what it wants to do - this is exactly what you observe. The compiler is not even comparing the numbers.

.LC0:
        .string "%d\n"
main:
        push    rax
        mov     esi, -2147483648
        mov     edi, OFFSET FLAT:.LC0
        xor     eax, eax
        call    printf
        mov     esi, 1
        mov     edi, OFFSET FLAT:.LC0
        xor     eax, eax
        call    printf
        xor     eax, eax
        pop     rdx
        ret

UB meand that the program behaviour is undefined ie it can be predicted from the C point of view.

Using GCC you can force the arithmetic and see what your hardware will do with it (but it is still undefined in C language):

int main()
{
    volatile int a=INT_MAX,b=1;
    printf("%d\n",a+b);
    printf("%d\n",(a+b)>a);

    return 0;
}

https://godbolt.org/z/6fjh8obfY