c - 是否在同一表达式未定义行为中两次调用具有局部副作用的函数？

Question

int f() {
    static int i=0;
    return ++i;
}
int g() {
    return f() + f();
}

g() 返回 3 还是结果 undefined？

Answer 1

Chapter and verse :

6.5.2.2 Function calls
...
10     There is a sequence point after the evaluations of the function designator and the actualarguments but before the actual call. Every evaluation in the calling function (includingother function calls) that is not otherwise specifically sequenced before or after theexecution of the body of the called function is indeterminately sequenced with respect tothe execution of the called function. ⁹⁴⁾

---

^{94) In other words, function executions do not ‘‘interleave’’ with each other}

结果是每个 ++i 之间都有一个序列点，因为它是函数调用的一部分。因此，此行为是明确定义的。

它是否真的按照您的意图进行是另一回事。请注意，在某些时候，您可能会面临未定义的签名溢出风险。正如其他人指出的那样，f() - f() 可能不会给出您期望的结果(在这种情况下不能保证从左到右的评估)。