What does the () operator do in C++

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Hi, I was reading my C++ book which a gave a very little information about the () operator - He said that's not very important, But I'm curious to know what is this operator, maybe I'll use it someday.

Thanks, Sam

Posted 9-Jul-15 15:33pm

Samuel Shokry

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Mohibur Rashid 9-Jul-15 21:55pm

keep reading the book

Samuel Shokry 9-Jul-15 22:20pm

Actually I have finished reading it.

Samuel Shokry 9-Jul-15 22:23pm

It is Practical C++ programming by Steve Oualline. He wrote exactly 3 lines about it plus 7 lines of simple example that I haven't understand well.

3 solutions

Solution 3

This gets to the point.
https://en.wikipedia.org/wiki/Function_object[^]

Posted 9-Jul-15 22:21pm

User 59241

Solution 1

This is a functional form or function-call operator or conversion (C-style cast) operator. I have no idea how a book can explain it all in 7 lines, and I have no enthusiasm to find out. Use something decent:
https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B[^],
http://www.cplusplus.com/doc/tutorial/operators[^],
http://www.tutorialspoint.com/cplusplus/function_call_operator_overloading.htm[^],
http://www.cplusplus.com/doc/tutorial/typecasting[^],
https://msdn.microsoft.com/en-us/library/df74sak1.aspx[^],
https://msdn.microsoft.com/en-us/library/ts48df3y.aspx[^].

—SA

Posted 9-Jul-15 17:47pm

Sergey Alexandrovich Kryukov

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Stefan_Lang 10-Jul-15 3:57am

I disagree about C-style cast operator - I'm sure you'll agree if you think about it for a moment. ;-)

Sergey Alexandrovich Kryukov 10-Jul-15 10:13am

I just mean: https://msdn.microsoft.com/en-us/library/ts48df3y.aspx (referenced above). Please, what are you disagree with?
Consider it is not "()", or what?
—SA

Andreas Gieriet 10-Jul-15 15:48pm

I agree with Stefan.
If talking about () operator, you usually mean the function call operator SomeTypeT operator()(some parameters ...) { ... }, I'd say.
The cast operator (if called is of the form (SomeTypeT)expr) is defined as operator SomeTypeT () { ... }.
Regards
Andi

Sergey Alexandrovich Kryukov 10-Jul-15 19:39pm

You are right, of course, but where did I say anything against it?
The question is about '()', without naming it anyhow; I just added some information about cast operator which can be defined as
operator const char *()
I thought that, in a way, it has something to do with '()' :-)
Where are you disagree, exactly?
—SA

Andreas Gieriet 12-Jul-15 9:15am

What about the placement-new operator? E.g. for a memory pool void* operator new(std::size_t, Pool*) { ... } or alike and call it MyClass *obj = new (myPool) MyClass;?
Just since the operator has some parenthesis in the expression does not mean that it has to do with the () operator... ;-)
So, I disagree that the cast operator has anything to do with the () operator.
Cheers
Andi

Sergey Alexandrovich Kryukov 12-Jul-15 13:28pm

Understood.
—SA

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Stefan_Lang · Accepted Answer · 2015-07-09T22:03:00

You can think of the () operator as a classs method without a name, because that is (almost) how it looks when you invoke this operator: rather than

C++

myobject.foo();
myobject.bar(1, 3.13, "hello world");

the calls would look like this:

C++

myobject();
myobject(1, 3.13, "hello world");

Note that the '.' is missing; this is on purpose, because the idea is to use the class object like a function. The main use of this operator is in functional programming (remember, C++ is not an object oriented language, it is a general purpose language, and therefore suitable for any programming paradigm!), and in generic programming.