如何自动从函数类型推断返回类型？

发布于 2024-11-17 00:41:13 字数 1349 浏览 1 评论 0原文

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（或兼容的东西）的指针，例如：

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);

由于我正在包装的类的设置器采用以下形式：

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

我写了一个“转换”功能：

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，效果很好，但我想知道是否有一种方法可以使其变得更加“容易”（使用）。

您是否认为有可能得到类似的结果：

// 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（甚至解释如何专门化 boost::python 来处理我的具体情况）。

谢谢。

原文

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.

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同展鸳鸯锦 2024-11-24 00:41:13

实际上可以使用强制转换运算符来推断函数的返回类型 - 我对此进行了描述

这是简短的版本：

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

这里，编译器将推断 T 的值，从而推断 func 的返回类型。

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.

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