后缀的优先级++

Question

考虑这些代码：

int a = 5;
int b = a++;
cout << "b is " << b << endl; // b is 5

或：

int get_number(){
    int a = 5;
    return a++;
}
int main(){
    int b = get_number();
    cout << "b is " << b << endl; // b is 5
}

根据 this table 后缀 ++ 运算符的优先级高于 = 运算符，因此输出应为 b is 6。但输出是 b is 5。我们该如何解释？

Answer 1

内置类型的 postfix++ 运算符（和 postfix-- 运算符）的约定是 returns 以前的值 ，而不管发生。所以函数仍然是 returns 5，即使收到变化的变量被赋值 6 之后。

Answer 2

根据expr.post.incr，强调我的：

The value of a postfix ++ expression is the value of its operand. [ Note: The value obtained is a copy of the original value — end note ]

The value computation of the ++ expression is sequenced before the modification of the operand object. With respect to an indeterminately-sequenced function call, the operation of postfix ++ is a single evaluation

int a = 5;
int b = a++; // the value computation for a is
             // the non-modified / non-incremented value
             // which is 5

std::cout << "a is " << a << std::endl; // a is 6
std::cout << "b is " << b << std::endl; // b is 5

后缀的优先级++

Precedence of postfix ++

c++

operators