如何从函数类型自动推断返回类型? [英] How to automatically infer the return type from a function type?

问题描述

我正在使用boost::python创建C ++库的Python包装器.在某些时候，boost::python需要一个指向成员函数(或兼容的函数)的指针，例如:

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);

由于我要包装的类具有以下形式的设置器:

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

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

我编写了一个转换"功能:

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);
}

我可以这样使用:

class Foo
{
  public:

    Foo& bar(int value);
};

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

到目前为止，这种方法很好用，但是我想知道是否有一种方法可以使它更加容易"(使用).

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

您认为可以通过某种方式获得类似的东西吗?

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));

欢迎所有解决方案(甚至包括解释如何专门化boost::python来处理我的特定情况的解决方案).

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

谢谢.

推荐答案

返回类型不能在C ++中自动推导，并且您不能基于返回类型重载函数.

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

在C ++ 0x中，有一个有趣的新功能称为 decltype .

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

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