什么是cout的正确答案< c ++<< C;？ [英] What is the correct answer for cout << c++ << c;?

问题描述

最近在采访中有一个以下的客观类型问题。

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

int c = 0;
cout << c++ << c;

答案：

10

b。 01

c。未定义的行为

a. 10
b. 01
c. undefined behavior

我回答选择b，即输出将是01。

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

我被一个访问员告诉，正确的答案是选项c：undefined。

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 ++中的序列点的概念。对于以下语句，行为未定义：

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< c ++<<将 ostream.operator c调用两次 ostream.operator （c ++）及更高版本 ostream.operator< <（c）。

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

我也检查了VS2010编译器的结果，其输出也是'01'。

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

推荐答案

>

You can think of:

cout<<c++<<c;

As：

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

C ++保证以前评估的所有副作用都将在序列点。在函数参数求值之间没有序列点，这意味着可以在参数 std :: operator<<<（std ::）之前计算参数 c cout，c ++）或之后。所以上面的结果是未定义的。

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 c can be evaluated before argument std::operator<<(std::cout, c++) or after. So the result of the above is undefined.

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