CODEWITHSUNDEEP
c++pointersoperator-overloading
Kvothe
Kvothe

Reputation: 1837

Combining member access operators with overloaded call operator

Given a class Foo with overloaded Boo* operator()(unsigned int), how should the overloaded operator be accessed?

I originally tried

std::unique_ptr<Foo> foo_ptr(new Foo);
Boo* boo_ptr = foo_ptr(1);

But this doesn't work, so I tried:

std::unique_ptr<Foo> foo_ptr(new Foo);
Boo* boo_ptr = foo_ptr->(1);

but this doesn't work either (and I didn't really expect it to). Instead I have to do

std::unique_ptr<Foo> foo_ptr(new Foo);
Boo* boo_ptr = (*foo_ptr)(1);

Which is more verbose than declaring a method in Foo such as ByIndex(unsigned int) instead (and making it shorter and more succinct was why I overloaded the operator in the first place).

Is there a way to do this without having to use the * operator to get the base value of the pointer?

Upvotes: 0

Views: 47

Answers (2)

Cornstalks
Cornstalks

Reputation: 38218

One way is to make a reference:

std::unique_ptr<Foo> foo_ptr(new Foo);
Foo& foo_ref = *foo_ptr;

Boo* boo_ptr = foo_ref(1);

Besides that, I think Mike Seymour is correct that there's nothing else you can do to invoke operator().

Upvotes: 2

Mike Seymour
Mike Seymour

Reputation: 254471

No, you have to dereference the pointer to access the object; so the only ways to call the operator are

(*foo_ptr)(1)
foo_ptr->operator()(1);

Upvotes: 4

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