Does std::endl work with both cout and wcout?

Question

I am wondering if std::endl works with both std::cout and std::wcout? Anyone is clear on this?

Accepted Answer

Yes. In fact, `std::endl` is a function template that will work as a manipulator on any specialization of the `std::basic_ostream` template. Some more detail: 27.7.3.6 prescribes that the `std::basic_ostream` template contain overload for `operator<<` as follows: ``` basic_ostream & operator<<(basic_ostream & (*pf)(basic_ostream &)); ``` The effect of invoking this overload on a suitable function is `return pf(*this)`. So when you say `std::cout << std::endl`, this actually becomes `std::endl(std::cout)` and returns a reference to the stream object. All other ostream manipulators are written in the same way, and similarly for input manipulators. The magic of the `endl` function template is a call to `widen(' ')`, which produces the correct "newline" data for the given character type.

Does std::endl work with both cout and wcout?

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