I am wondering if std::endl works with both std::cout and std::wcout?
Anyone is clear on this?
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How about writing the code to test it ? – asheeshr Dec 01 '12 at 13:18
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I assume he means to ask if there is anything wrong with him using endl with wcout. – ctor Dec 01 '12 at 13:19
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I added some more detail explaining how manipulators work. – Kerrek SB Dec 01 '12 at 14:10
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Yes. In fact, std::endl 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<charT, traits> &
operator<<(basic_ostream<charT, traits> (*pf)(basic_ostream<charT, traits> &));
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('\n'), which produces the correct "newline" data for the given character type.
Kerrek SB
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