C ++ template：获取没有对象的成员函数的返回类型 [英] C++ template: get return type of member function without an object

问题描述

我有一些类，我不能修改。每个都有一个复制构造函数，至少一个其他构造函数和一个返回一些值的函数 foo（）。我想创建一个类模板，可以从这些类派生，并有一个数据成员与 foo（）（对不起if我有一些术语错了。）

I have a number of classes that I cannot modify. Each has a copy constructor, at least one other constructor, and a function foo() that returns some value. I want to make a class template that can derive from each of these classes, and has a data member that is the same type as the return type of foo() (sorry if I've got some of the terminology wrong).

换句话说，我想要一个类模板

In other words, I would like a class template

template<typename T> class C : public T
{
  footype fooresult;
};

其中 footype code> T :: foo（）。

where footype is the return type of T::foo().

如果基类都有一个默认的构造函数，

If the base classes all had, say, a default constructor, I could do

decltype(T().foo()) fooresult;

（在GCC中使用C ++ 0x功能），但类没有任何特定的构造函数

(with the C++0x functionality in GCC) but the classes don't have any particular constructor in common, apart from the copy constructors.

GCC也不允许 decltype（this-> foo（）），虽然显然有一个可能性，这将被添加到C ++ 0x标准 - 任何人都知道这是多少可能是？

GCC also doesn't allow decltype(this->foo()), though apparently there is a possibility that this will be added to the C++0x standard - does anyone know how likely that is?

我觉得应该可以按照 decltype（foo（））或 decltype（T :: foo（））但是那些似乎不工作：GCC给出的形式的错误不能调用成员函数'int A :: foo（）'没有对象。

I feel like it should be possible to do something along the lines of decltype(foo()) or decltype(T::foo()) but those don't seem to work: GCC gives an error of the form cannot call member function 'int A::foo()' without object.

当然，我可以有一个额外的模板参数 footype ，或者甚至类型<$ c $的非类参数

Of course, I could have an extra template parameter footype, or even a non-class parameter of type T, but is there any way of avoiding this?

推荐答案

你不需要记住，因为decltype不会评估它的参数，你可以调用 nullptr 。

You don't need that- remember that since decltype doesn't evaluate its argument, you can just call on nullptr.

decltype(((T*)nullptr)->foo()) footype;

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