I am new to C++ and I am confused about std::endl. When trying to understand what std::endl is, I had faced some resources which told me that it is a function.
However, how can a function be deprived of parentheses?
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5There's a difference between a function and a function call. – molbdnilo Sep 21 '17 at 11:24
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1A function without parentheses is like an integer without a `+`, or an aray without a `[]`, or a string without a `printf`. You pass it around, put it aside, then when its time comes, you retrieve it and stick it in front of a list of arguments. (Technically you are oassing a function pointer or reference around, but this is not important at this moment). – n. m. could be an AI Sep 21 '17 at 11:36
3 Answers
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Read the ref:
std::endlInserts a new-line character and flushes the stream.
It's used with a stream, for example std::cout.
It's not a function, it's a function template.
std::endl without parentheses refers to a set of overload functions - all possible specializations of that function template. Read more in How does std::endl not use any brackets if it is a function?
gsamaras
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Even being a function template, it is still a function and OP is asking why you dont need to call it as `std::endl()` – Amadeus Sep 21 '17 at 11:27
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However, how can a function be deprived of parentheses?
The name of a function, without being followed by (), is just a reference to that function. It's exactly the same as with any other type:
void foo(int) {}
char x = 'a';
char *p = &x;
int main()
{
p; // Refers to p
*p; // Dereferences p (refers to whatever p points to)
foo; // Refers to foo
foo(42); // Calls foo
}
std::endl is a function (actually a function template) which takes one parameter of type "a stream", and works by inserting an EOL representation into that stream and then flushing it. You can actually use it like any other function, if you want to:
std::endl(std::cout);
The final piece of the puzzle is that the standard library provides an overload (again, a template) of operator << such that the LHS argument is a stream and the RHS argument is a function; the implementation of this operator calls the RHS argument (the function) and passes it the LHS one (the stream). Conceptually, there's something like this:
Stream& operator<< (Stream &s, const Function &f)
{
f(s);
return s;
}
Therefore, calling std::cout << std::endl invokes that operator overload, which in turn invokes std::endl(std::cout), which does the EOL insertion + flushing.
As to which form is to be preferred (direct call vs. << operator), it's definitely the use of <<. It's idiomatic, and it allows easy composition of multiple stream manipulators within a single expression. Like this:
std::cout << "Temperature: " << std::fixed << std::setprecision(3) << temperature << " (rounds to " << std::setprecision(1) << temperature << ')' << std::endl;
Angew is no longer proud of SO
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Calling `std::cout << std::endl(std::cout)` gives an error, or am I doing something wrong? – Arrrow Sep 21 '17 at 11:41
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1@Arrrow Try `std::endl(std::cout);` instead of `std::cout << std::endl;` as a single statement. – Bob__ Sep 21 '17 at 11:42
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2@Arrrow you are trying to send return value of `std::endl(std::cout)` function to stream `std::cout`, and that's stream reference, so you are doing something like `std::cout << std::cout`, with intermission of sending also EOL to the stream by the `endl` function. I.e. doesn't make that much sense. – Ped7g Sep 21 '17 at 11:44
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@Bob__ Ah so this has the same functionality as std::cout << std::endl? – Arrrow Sep 21 '17 at 11:48
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@Arrrow I never spoke about streaming the result of calling `std::endl(std::cout)` anywhere. Just call it like a plain old function. – Angew is no longer proud of SO Sep 21 '17 at 11:49
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@Arrrow Yes, that's what the answer says. Doing `std::cout << std::endl` is effectively syntactic sugar for calling `std::endl(std::cout)`, which inserts the newline & flushes the stream. – Angew is no longer proud of SO Sep 21 '17 at 11:50
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@Angew Is there a reason to use one above the other? I always learned to do `std::cout << std::endl` at least. – Arrrow Sep 21 '17 at 11:52
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@Arrrow Of course, `std::cout << std::endl` is indeed the idiomatic way. I would never write `std::endl(std::cout)` in real code, since it's unfamiliar. I was just explaining how it works under the hood. – Angew is no longer proud of SO Sep 21 '17 at 11:55
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It should be noted that the reason to use stream manipulators at all is that they can be easily composed with ``operator <<``, e.g. ``std::cout << "The temperature is " << std::fixed << std::right << std::setw(5) << std::setprecision(2) << temperature << std::endl;``; For direct function calls, you'd always need to use multiple statements or ``operator ,`` to avoid ordering issues. – Arne Vogel Sep 21 '17 at 13:49
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@ArneVogel Good point, thanks. I'll incorporate it into the answer when I have the time (probably tomorrow). – Angew is no longer proud of SO Sep 21 '17 at 13:56
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endl is an output-only I/O manipulator.
endl is an output-only I/O manipulator, it may be called with an expression such as
out << std::endlfor any out of typestd::basic_ostream.Inserts a newline character into the output sequence os and flushes it as if by calling
os.put(os.widen('\n'))followed byos.flush().
