How to automatically infer the return type from a function type?

问题

I'm using boost::python to create a Python wrapper of a C++ library. At some point, boost::python needs a pointer to a member function (or something compatible), like:

template <class MyClass, typename ValueType>
void (*setter_function)(MyClass&, ValueType)

// This doesn't compile, but you got the idea.
my_boost_python_call(setter_function f);

Since the class I'm wrapping has its setter in the following form:

template <class MyClass, typename ValueType>
MyClass& (MyClass::*setter_method)(ValueType)

I wrote a single "conversion" function:

template <typename MyClass, typename ValueType, setter_method<MyClass, ValueType> fluent_setter>
void nonfluent_setter(MyClass& instance, ValueType value)
{
  (instance.*fluent_setter)(value);
}

Which I can use like that:

class Foo
{
  public:

    Foo& bar(int value);
};

my_boost_python_call(nonfluent_setter<Foo, int, &Foo::bar>);

So far this works well, but I wonder if there is a way to make this even more "easy" (to use).

Do you think it is somehow possible to get something like:

// return type is somehow inferred from the member function pointer
my_boost_python_call(nonfluent_setter<Foo, &Foo::bar>);

// or even a syntax similar to boost::python::make_function
my_boost_python_call(make_nonfluent_setter<Foo>(&Foo::bar));

All solutions (even ones that explain how to specialize boost::python to handle my specific case) are welcome.

Thank you.

回答1:

The return type cannot be automatically deduced in C++ and you cannot overload functions based on return type.

In C++0x there is a new feature called decltype that might be of interest.

回答2:

It is actually possible to infer the return type of a function using casting operators -- I describe this here.

Here's the short version:

struct ReturnType {
  template<typename T>
  operator T() {
    // your code for return a T goes here.
  }
};
ReturnType func() { return ReturnType(); }

Here, the compiler will infer the value of T, thus the return type of func is also inferred.

来源：https://stackoverflow.com/questions/6439769/how-to-automatically-infer-the-return-type-from-a-function-type

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c++

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metaprogramming

boost-python