std::cout 有返回值吗?
我很好奇 std::cout 是否有返回值,因为当我这样做时:
I am curious if std::cout has a return value, because when I do this:
cout << cout << "";
打印了一些十六进制代码.这个打印出来的值是什么意思?
some hexa code is printed. What's the meaning of this printed value?
推荐答案
因为cout的操作数<<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 operator<<
,在这种情况下,它们都是 std::ostream
类型.
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
.
有许多候选运算符重载可供选择,但我将按照通常的重载解决过程来描述最终被选中的那个.
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
有一个转换运算符,允许转换为 void*
.这用于将流的状态测试为布尔条件(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).
右侧操作数表达式 cout
使用此转换运算符隐式转换为 void const*
,然后 operator<<<
重载需要一个 ostream&
和一个 void const*
来写入这个指针值.
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,从 ostream
到 void*
转换产生的实际值是未指定的.规范仅要求如果流处于错误状态,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 operator<<
重载确实有一个返回值:它们返回作为操作数提供的流.这就是允许插入操作链接的原因(对于输入流,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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