C output changed by adding a printf

Question

int32_t result = registers[rs] + registers[rt];  
if (((registers[rs] > 0) && (registers[rt] > 0) && (result < 0)) || 
    ((registers[rs] < 0) && (registers[rt] < 0) && (result > 0))) {
        fprintf(stderr, "arithmetic overflow\n");
} else {
        registers[rd] = result;
}

Just a simple add and detect the overflow, if registers[rs] is 2147483647 and registers[rt] is 1, registers[rd] will have a result -2147483648. But if I add a printf("%d\\n", result) after declaring the variable result with same value in rs and rt, it will display arithmetic overflow

when registers[rs] is 2147483647 and registers[rt] is 1, it should output the arithmetic overflow

Answer 1

int32_t result = registers[rs] + registers[rt];  

When signed integer overflow occurs it invokes UB(Undefined Bahaviour).

You need to check before you add those integers.

bool willOverflow(int32_t a, int32_t b) 
{
    bool overflow = false;

    if (a > 0 && b > 0 && (a > INT32_MAX - b))  
    {
        overflow = true;
    }
    else if (a < 0 && b < 0 && (a < INT32_MIN - b)) 
    {
        overflow = true;
    }

    return overflow;
}

if (willOverflow(registers[rs], registers[rt]))
{
    fprintf(stderr, "arithmetic overflow\n");
} 
else 
{
    registers[rd] = registers[rs] + registers[rt];
}