I am wondering if std::endl works with both std::cout and std::wcout?
Anyone is clear on this?
I am wondering if std::endl works with both std::cout and std::wcout? Anyone is clear
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I am wondering if std::endl works with both std::cout and std::wcout?
Anyone is clear on this?
Yes. In fact,
std::endlis a function template that will work as a manipulator on any specialization of thestd::basic_ostreamtemplate.Some more detail: 27.7.3.6 prescribes that the
std::basic_ostreamtemplate contain overload foroperator<<as follows:The effect of invoking this overload on a suitable function is
return pf(*this). So when you saystd::cout << std::endl, this actually becomesstd::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
endlfunction template is a call towiden('\n'), which produces the correct "newline" data for the given character type.