获取没有对象的成员函数的返回类型 [英] Get return type of member function without an object
问题描述
我有许多无法修改的类。每个函数都有一个副本构造函数，至少一个其他构造函数和一个返回某些值的函数 foo（）。我想制作一个可以从这些类的每一个派生的类模板，并具有与 foo（）的返回类型相同的数据成员类型（对不起，如果我的某些术语有误）。
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 是 T :: foo（）。
如果所有基类都具有默认构造函数，则可以
If the base classes all had, say, a default constructor, I could do
decltype(T().foo()) fooresult;
（在GCC中具有C ++ 0x功能），但是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（）），尽管显然有可能将其添加到C ++ 0x标准中-有人知道这有多大可能性吗？
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?
应该可以按照 decltype（foo（））或 decltype（T :: foo（））但那些似乎不起作用：GCC给出了错误形式，无法在没有对象的情况下调用成员函数'int A :: foo（）'。
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 ，甚至还有一个类型为<$ c $的非类参数。 c> T ，但是有什么方法可以避免这种情况？
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
您不需要那-请记住，由于decltype不评估其参数，因此您可以调用 nu llptr 。
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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