Question

I was reading about techniques to detect overflow in C . one of the examples to show incorrect solution to detect overflow in addition was this one :

/* Determine whether arguments can be added without overflow */
int tadd_ok(int x, int y) {
    int sum = x+y;
    return (sum-x == y) && (sum-y == x);
}

and it said it doesn’t work because :

two’s-complement addition forms an abelian group, and so the
expression (x+y)-x will evaluate to y regardless of whether or not the
addition overflows, and that (x+y)-y will always evaluate to x

What does it exactly mean ? Does it mean that C compiler replace sum with x+y ?
To figure out what is it saying I even traced assembly code of the program, but there was no sign of replacement .

Update : The essence of my question is, does GCC evaluates an expression without calculating it ?
This is NOT a question about two’s complement.
You can see a sample output in here .

Answer 1

If you take a trivial example of 4 (0b0100) + 5 (0b0101) you can see that the unsigned sum should be 9 (1001) which is actually -7 in two’s complement. If you then take that sum (0b1001) and subtract 4 from it using two’s complement arithmetic:

    0b1001 - 0b0100 = 0b1001 + 2s_complement(0b0100) = 0b1001 + 0b1100 = 0b1_0101 

you end up with 0101 which is 5 (you drop the overflowing most significant 1 during a 2’s complement operation). Subtracting 5 from the sum equals 4:

    0b1001 - 0b0101 = 0b1001 + 2s_complement(0b0101) = 0b1001 + 0b1011 = 0b1_0100

This satisfies the c code you provided but still resulted in an overflow.

From wikipedia’s article on two’s complement:

Two's complement    Decimal
0111                 7
0110                 6
0101                 5
0100                 4
0011                 3
0010                 2
0001                 1
0000                 0
1111                −1
1110                −2
1101                −3
1100                −4
1011                −5
1010                −6
1001                −7
1000                −8

Update:
To demonstrate your INT_MAX example using my trivial 4 bit integer system with INT_MAX = 7 we can see the same result as your c code.

    7 + 7 (0b0111 + 0b0111) = 0b1110 (-2 in two's complement)

Just like my example above, subtracting, sum - 7 will equal 7.

    0b1110 - 0b0111 = 0b1110 + 2s_complement(0b0111) = 0b1110 + 0b1001 = 0b1_0111