what is Base()()

Question

Can someone please explain the Base*(*)() for me as in:

typedef std::map<std::string, Base*(*)()> map_type;

And how would one return it from a function?

I presume it is a function pointer, a Base* is returned, but what is this (*).

I found this in the following SO post Is there a way to instantiate objects from a string holding their class name?

Thanks

[cdecl](http://cdecl.org/) says `Base*(*)()` is a syntax error, interesting. — fredoverflow, Aug 05 '12 at 14:12
@FredOverflow: I think we have to assume that `Base` is defined as a type. — CB Bailey, Aug 05 '12 at 14:16

CB Bailey · Accepted Answer · 2012-08-05T13:58:26.123

7

Base* (*)() is a type: pointer to function returning Base*. The * means that it is a pointer and the () are used to to override the precedence to ensure that the pointer applies to the function itself and not the return type.

You can return it from a function by returning the name of a function of the appropriate type.

E.g.

Base* f();

Base* (*g())()
{
    return f;
}

edited Aug 05 '12 at 13:58

answered Aug 05 '12 at 13:51

CB Bailey

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score 4 · Answer 2 · answered Aug 05 '12 at 13:52

4

It's the type of a function pointer of a function of signature Base*():

Base * foo();

Base * (*fp)() = &foo;

Or in your case:

map_type callbacks;
callbacks["Foo"] = &foo;

To invoke:

Base * p = callbacks["Foo"]();  // same as "p = foo();"

answered Aug 05 '12 at 13:52

Kerrek SB

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Thanks for your help, unfortunately I can't up-vote your answer due to a lack of points. – Steven ZA Aug 05 '12 at 15:28

what is Base*(*)()

2 Answers2

what is Base()()