std :: cout有返回值吗？ [英] Does std::cout have a return value?

问题描述

我很好奇，如果std :: cout有一个返回值，因为当我这样做：

I am curious if std::cout has a return value, because when I do this:

cout << cout << "";

打印一些六进制代码。

因为 cout<< p>的操作数，所以这个打印值的含义是什么？

some hexa code is printed. What's the meaning of this printed value?

推荐答案

cout 是用户定义的类型，表达式实际上是一个函数调用。编译器必须找到与操作数匹配的最佳运算符<< ，在这种情况下，操作数都是类型 std :: ostream 。

Because the operands of cout << cout are user-defined types, the expression is effectively a function call. The compiler must find the best operator<< that matches the operands, which in this case are both of type std::ostream.

有很多候选运算符重载可供选择，但我将描述最终选择的运算符重载，遵循通常的重载解决过程。

There are many candidate operator overloads from which to choose, but I'll just describe the one that ends up getting selected, following the usual overload resolution process.

std :: ostream 有一个转换操作符，允许转换为 void * 。这用于使能测试流的状态作为布尔条件（即，它允许 if（cout）工作）。

std::ostream has a conversion operator that allows conversion to void*. This is used to enable testing the state of the stream as a boolean condition (i.e., it allows if (cout) to work).

右侧的操作数表达式 cout 隐式转换为 void const * code>使用这个转换运算符，然后运算符<< 重载，它接受一个 ostream& void const * 被调用以写入此指针值。

The right-hand operand expression cout is implicitly converted to void const* using this conversion operator, then the operator<< overload that takes an ostream& and a void const* is called to write this pointer value.

请注意， $ c> ostream 到 void * 转换未指定。规范只要求如果流处于坏状态，则返回空指针，否则返回非空指针。

Note that the actual value resulting from the ostream to void* conversion is unspecified. The specification only mandates that if the stream is in a bad state, a null pointer is returned, otherwise a non-null pointer is returned.

用于流插入的运算符<< 重载有一个返回值：它们返回作为操作数提供的流。这是允许插入操作链接（以及输入流，使用>> 的提取操作）。

The operator<< overloads for stream insertion do have a return value: they return the stream that was provided as an operand. This is what allows chaining of insertion operations (and for input streams, extraction operations using >>).

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