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c++ syntax question

Posted on 2016-10-10
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Last Modified: 2016-10-10
I do not know how to read following statement:

    auto t = create_task([]() -> int
    {
        return 42;
    });

I can understand Lamda part of the code, but I could not understand "->int" part, what does it mean, how does it affecting the result? Thank you for any help.
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Question by:Evan Li
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Expert Comment

by:evilrix
ID: 41837271
It's a function that is being passed a C++14 Lambda function that returns an int, in this case the value 42.

More information on C++ lambda functions:

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Expert Comment

by:Pawan Kumar Khowal
ID: 41837272
-> This is the Arrow Operator. It is a dereference operator that is used exclusively with pointers to objects that have members. This operator serves to access the member of an object directly from its address.  For accessing object variables and methods via pointer to object.

If p_emp is a pointer to an object of type Employee, then to assign the value "zara" to the first_name member of object emp, you would write something as follows:

strcpy(p_emp->first_name, "zara");

Below will also explain you the meaning of :: ,->, : in C++?

Ref - https://www.quora.com/What-is-the-meaning-of-in-C++

I hope it helps.
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Expert Comment

by:evilrix
ID: 41837279
It could be re-written like this to make it a little clearer:

auto lambda = []() -> int { return 42; }; // lambda is just a variable that holds the lambda function
auto t = create_task(lambda); // this is now passing the function object to the create_task function

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Accepted Solution

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evilrix earned 500 total points
ID: 41837283
Pawan,

I'm afraid your assessment of the code is incorrect. In this case, the -> is not being used to dereference anything, it is the syntax used to specify the return value of an auto return value lambda. This syntax was introduce in C++14.
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Author Comment

by:Evan Li
ID: 41837285
OK, I found the answer, "->int" means that the lambda function will return a integer. Thanks for your attention.
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Author Closing Comment

by:Evan Li
ID: 41837288
You are right, it is the return type
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Expert Comment

by:evilrix
ID: 41837290
Indeed, but that's only in this case. That syntax is specifically used to allow the use of the decltype operator to be able to specify the return type of a lambda from one or more of the passed arguments. In this case, it's completely pointless, because the return type isn't being deferred from a passed argument.
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Expert Comment

by:Pawan Kumar Khowal
ID: 41837294
Thnx evilrix ! I got in pointers that why got confused. Thanx for the info.
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Expert Comment

by:evilrix
ID: 41837296
No worries, Pawan. Have a good day, sir.
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