How does std::endl not use any brackets if it is a function?

Question

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?

Although not a duplicate question, the answers here answer your question: [4842901](http://stackoverflow.com/questions/4842901/c-statement-cannot-resolve-address-for-overloaded-function). — CB Bailey, Oct 02 '11 at 00:11

CB Bailey · Accepted Answer · 2011-10-02T00:27:07.737

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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.

edited Oct 02 '11 at 00:27

answered Oct 02 '11 at 00:19

CB Bailey

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Thanks for this explanation! Is there a way of passing std::endl to a class, like in the following: http://stackoverflow.com/q/39019763/484230 – Martin Aug 18 '16 at 13:42
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For those wondering, yes, you can do `std::ostream & some_func(std::ostream &stream) { return stream << ":D"; }` and use it by doing `std::cout << some_func;`. – Qix - MONICA WAS MISTREATED Nov 13 '16 at 05:56

Flame · Answer 2 · 2011-10-02T06:53:42.667

1

std::endl is an object of some type (not really important) that is supplied as an argument to operator<<( std::ostream &, decltype(std::endl)).

EDIT

Reading the other question would lead me to think that endl is a function template and that we most likely select the ostream& operator<<(ostream&(*)(ostream&)) member function overload.

edited Oct 02 '11 at 06:53

answered Oct 02 '11 at 00:12

Flame

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`std::endl` isn't an object (in the C++ sense), it's a function template. – CB Bailey Oct 02 '11 at 00:15
... which is why there's no `std::wendl`. I'm not sure if that's a good thing in hindsight. – MSalters Oct 02 '11 at 01:01
@MSalters: I'm not sure how that's a reason. – Lightness Races in Orbit Oct 02 '11 at 01:09
@Tomalak : Well, there's `'\n'` and a distinct `L'\n'`. If you're thinking of `endl` as some object that's like `'\n'`, you'd expect a `wendl`. – MSalters Oct 02 '11 at 01:14
@MSalters: I agree! Which is why I don't see how that `endl` is _able_ to be both is a reason that it _does_. – Lightness Races in Orbit Oct 02 '11 at 11:17

Lightness Races in Orbit · Answer 3 · 2011-10-02T01:09:37.587

1

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()).

edited Oct 02 '11 at 01:09

answered Oct 02 '11 at 00:28

Lightness Races in Orbit

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**Note:** My phone really needs a backtick key. – Lightness Races in Orbit Oct 02 '11 at 00:32
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`std::endl` isn't a class with an `operator()`; it's a function template. – CB Bailey Oct 02 '11 at 00:45

How does std::endl not use any brackets if it is a function?

3 Answers3

EDIT

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