Menu
The question is pretty much in the title. According to C++ Reference, std::endl is actually a function. Looking at its declaration in <iostream>, this can be verified.
However, when you use std::endl, you don't use std::endl(). Instead, you use:
std::cout << "Foo" << std::endl; In fact, if you use std::endl(), the compiler demands more parameters, as noted on the link above.
Would someone care to explain this? What is so special about std::endl? Can we implement functions that do not require any brackets when calling too?
891
C++ manipulator endl function is used to insert a new line character and flush the stream. Working of endl manipulator is similar to '\n' character in C++.
std::endl. Inserts a new-line character and flushes the stream. Its behavior is equivalent to calling os.
There is rarely a need for std::endl , and getting in the habit of using it will lead to mysteriously slow code. Just use '\n' unless you know you need to flush the buffer.
Both endl and \n serve the same purpose in C++ – they insert a new line. However, the key difference between them is that endl causes a flushing of the output buffer every time it is called, whereas \n does not.
std::endl is a function template declared (27.7.3.8):
template <class charT, class traits> basic_ostream<charT,traits>& endl(basic_ostream<charT,traits>& os); The reason that you can "stream" it to std::cout is that the basic_ostream class template has a member declared:
basic_ostream<charT,traits>& operator<< ( basic_ostream<charT,traits>& (*pf)(basic_ostream<charT,traits>&) ); which is defined to have the effect of returning pf(*this) (27.7.3.6.3).
std::endl without parentheses refers to a set of overload functions - all possible specializations of the function template, but used in a context where a function pointer of one particular type is acceptable (i.e. as an argument to operator<<), the correct specialization can be unambiguously deduced.
196
Though it's a function [template], standard stream manipulators are designed to be sent to streams as function pointers (or functor object references). Inserting the result of a function call won't give you anything but the value that results from that function call.
This means that you stream the functor itself (f), rather than the result of calling it (f()).
41
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
