这个检测整数加法溢出的函数实际上有用吗？

Question

While reading the comments for this question , I came across a link to the comp.lang.c FAQ that shows a "careful addition function" which purportedly detects integer overflow:

int
chkadd(int a, int b)
{
    if (INT_MAX - b < a) {
        fputs("int overflow\n", stderr);
        return INT_MAX;
    }
    return a + b;
}

How does this not overflow if b == -1 ? If the assumption is that a and b are both positive, why make them int rather than unsigned int in the first place?

Answer 1

OP has identified that INT_MAX - b may overflow, rendering the remaining code invalid for proper overflow detection. It does not work.

if (INT_MAX - b < a) {  // Invalid overflow detection

A method to detect overflow without UB follows:

int is_undefined_add1(int a, int b) {
  return (a < 0) ? (b < INT_MIN - a) : (b > INT_MAX - a);
}

---

why make them int rather than unsigned int in the first place?

Changing to unsigned does not solve the problem in general . The range of unsigned: [0...UINT_MAX] could be half of that of int: [INT_MIN...INT_MAX] . IOWs: INT_MAX == UINT_MAX . Such systems are rare these days. IAC, changing types is not needed as coded with is_undefined_add1() .

Answer 2

Probably they just overlooked it. Additional links on the FAQ page seem to provide more correct code.