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template <typename T>
class smart_ptr
{
public:
    // ... removed other member functions for simplicity
    T* get() { return ptr; }

    template <typename U>
    auto operator [](U u) const -> decltype((*get())[u])
    {
        return (*get())[u];
    }

    template <typename U>
    auto operator [](U u) -> decltype((*get())[u])
    {
        return (*get())[u];
    }

/*
    // These work fine:

    template <typename U>
    int operator [](U u)
    {
        return (*get())[u];
    }

    template <typename U>
    int& operator [](U u)
    {
        return (*get())[u];
    }
*/
private:
    T* ptr;
};

struct Test
{
};

struct Test2
{
    int& operator [](int i) { return m_Val; }
    int operator [](int i) const { return m_Val; }

    int m_Val;
};

int main()
{
    smart_ptr<Test> p1;
    smart_ptr<Test2> p2;
    p2[0] = 1;
}

错误:

prog.cpp: In function 'int main()':
prog.cpp:55:9: error: no match for 'operator[]' in 'p2[0]'

ideone:http: //ideone.com/VyjJ28

我试图在没有明确指定operator []返回类型的情况下使 smart_ptr 与返回类型一起工作。但是,从上面很明显,编译器无法编译代码。如果有人可以帮助我,我将不胜感激。T::operator []int

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1 回答 1

2

看来您想要更好的编译器错误。以下是clang对这个来源的看法(好吧,它说了很多,但这描述了问题):

decltype.cpp: In instantiation of ‘class smart_ptr<Test>’:
decltype.cpp:53:21:   required from here
decltype.cpp:9:51:error: cannot call member function ‘T* smart_ptr<T>::get() [with T = Test]’ without object
     auto operator [](U u) const -> decltype((*get())[u])       
                                                   ^

该问题的解决方法是调用get()一个对象,例如:

auto operator[](U u) const -> decltype((*this->get())[u])

(当然,这也要求有一个const成员get())。

于 2012-11-28T01:23:46.610 回答