模板类中的Typedef不起作用 [英] Typedef inside template class doesn't work

问题描述

我对以下代码有疑问:

template <typename U>
class lamePtr
{
public:
    typedef U* ptr;
};

template <typename U>
class smarterPointer
{
    public:
    void funFun()
    {
        typedef lamePtr<U> someType;
        someType::ptr query;
    }
};

如您所见，我在lamePtr中有一个typedef.在smarterPointer类内部，我有一个函数funFun().我想做的是使另一个typedef someType.直到那一行，一切正常为止，直到我们通过someType :: ptr查询到达该行为止.

As you see, I have a typedef inside lamePtr. Inside smarterPointer class I have a function funFun(). What am I trying to do is to make another typedef someType. Till that line, everything works fine until we get to the line with someType::ptr query.

我想在这里发生的是查询"将变为lamePtr< U> :: ptr(一个简单值，不是typedef;).但是，出现编译错误(使用gcc 4.4.3):

What I want here to happen is that "query" will become lamePtr< U >::ptr (a simple value, not a typedef ;). However, I get compilation errors (with gcc 4.4.3):

temp.cpp: In member function ‘void smarterPointer<U>::funFun()’:
temp.cpp:15: error: expected ‘;’ before ‘query’

我在做什么错了?

推荐答案

someType，因为lamePtr<U>是从属名称".是否有成员ptr取决于U是什么，如果有，则取决于该成员是什么样的事物".

someType, as lamePtr<U> is a "dependant name". It depends on what U is as to whether or not there is a member ptr and, if so, what kind of "thing" that member is.

当然，您知道对于所有T，lamePtr<T>::ptr是一种类型，但是在编译的这一阶段，解析器并不知道这一点.

Of course, you know that for all T, lamePtr<T>::ptr is a type, but at this stage of compilation the parser does not know that.

使用 typename关键字提示解析器一种.其余的将在以后的编译过程中解决.只是一点C ++怪癖.

Use the typename keyword to hint to the parser that it's a type. The rest will be resolved later in the compilation process. Just a little C++ quirk.

template <typename U>
class lamePtr
{
public:
    typedef U* ptr;
};

template <typename U>
class smarterPointer
{
    public:
    void funFun()
    {
        typedef lamePtr<U> someType;
        typename someType::ptr query;
    }
};

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