CODEWITHSUNDEEP
c++templatespointers
Jack D.
Jack D.

Reputation: 33

Making a function template with cv-qualified arguments of this function

I have a function template:

template < typename A, typename B, typename xClass, typename D>
void foo(A& a, B& (xClass::*fooPtr)(D& d));
{
//some code
}

I have some classes, which I want to use later as an xClass in my function. But their member functions have different const-qualifiers:

const out_t& Class1::foo1(in_t) const // (1)
out_t& Class1::foo1(in_t) // (2) just non-const version of foo1(), making the same actions
out_t& Class2::foo2(in_t) const // (3)

Now, if I use

... B& (xClass::*fooPtr)(D& d) ...

in function template, only pointer to (2) member function makes foo() work, others cause "incompatible cv-qualifiers" error. Howewer, adding "const" qualifier:

... B& (xClass::*fooPtr)(D& d) const ...

makes (3) working, but now (2) (and (1) also) causes error.

Is there a simple way to make function foo() to "do not notice" cv-qualifiers of pointers to member functions? I tried to use reinterpret_cast in order to make all necessary pointers to member functions similarly cv-qualified:

typedef B& out_t& (Class1::*fooPtr)(in_t) const;
fooPtr myPtr = reinterpret_cast<fooPtr>(&Class1::foo1);

but it failed.

Upvotes: 1

Views: 466

Answers (1)

Vaughn Cato
Vaughn Cato

Reputation: 64308

One approach is to make your foo function more general, so that it just takes a member pointer of any type:

template < typename A, typename F, typename xClass>
void foo(A& a, F xClass::*fooPtr)
{
//some code
}

This means that foo will accept pointers to data members or member functions that have signatures other than what you want, but it usually isn't that big of a problem. If you pass the wrong type of member, you'll typically just get a compilation error because the member function can't be called the way you are calling it. If you want to restrict it more, you can use SFINAE tricks.

Upvotes: 1

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