cout<<的正确答案是什么？ a ++<<一个;？ [英] What is the correct answer for cout << a++ << a;?

问题描述

最近在一次采访中出现了以下客观类型问题。

Recently in an interview there was a following objective type question.

int a = 0;
cout << a++ << a;

答案：

a。 10

b。 01

c。未定义的行为

a. 10
b. 01
c. undefined behavior

我回答了选项b，即输出为 01。

I answered choice b, i.e. output would be "01".

但后来令我惊讶的是一名面试官告诉我，正确的答案是选项c：未定义。

But to my surprise later I was told by an interviewer that the correct answer is option c: undefined.

现在，我知道C ++中序列点的概念。对于以下语句，该行为是未定义的：

Now, I do know the concept of sequence points in C++. The behavior is undefined for the following statement:

int i = 0;
i += i++ + i++;

但根据我对语句 cout<<的理解， a ++<< a 中， ostream.operator<（<（）将被调用两次，首先是 ostream.operator<< （a ++）和更高版本的 ostream.operator<<（a）。

but as per my understanding for the statement cout << a++ << a , the ostream.operator<<() would be called twice, first with ostream.operator<<(a++) and later ostream.operator<<(a).

我还在VS2010编译器上检查了结果，其输出也为'01'。

I also checked the result on VS2010 compiler and its output is also '01'.

推荐答案

您可以想到：

cout << a++ << a;

作为：

std::operator<<(std::operator<<(std::cout, a++), a);

C ++保证先前评估的所有副作用都将在序列点。函数参数评估之间没有顺序点，这意味着参数 a 可以在参数 std :: operator<<（std :: cout，a ++）或之后。因此，上面的结果是不确定的。

C++ guarantees that all side effects of previous evaluations will have been performed at sequence points. There are no sequence points in between function arguments evaluation which means that argument a can be evaluated before argument std::operator<<(std::cout, a++) or after. So the result of the above is undefined.

C ++ 17更新

在C ++ 17中，规则已更新。特别是：

In C++17 the rules have been updated. In particular:

在移位运算符表达式中 E1 << E2 和 E1>> E2 ， E1 的每个值计算和副作用都要在 E2 。

In a shift operator expression E1<<E2 and E1>>E2, every value computation and side-effect of E1 is sequenced before every value computation and side effect of E2.

这意味着它需要代码来生成结果 b ，它输出 01 。

Which means that it requires the code to produce result b, which outputs 01.

请参见 P0145R3完善惯用C ++的表达式评估顺序

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