C++ 模板类型名迭代器

问题描述

考虑以下头文件:

template <typename T> struct tNode
{
    T Data;                      //the data contained within this node
    list<tNode<T>*> SubNodes;       //a list of tNodes pointers under this tNode

    tNode(const T& theData)
    //PRE:  theData is initialized
    //POST: this->data == theData and this->SubNodes have an initial capacity
    //      equal to INIT_CAPACITY, it is set to the head of SubNodes
    {
        this->Data = theData;
        SubNodes(INIT_CAPACITY);   //INIT_CAPACITY is 10
    }

};

现在考虑来自另一个文件的一行代码:

Now consider a line of code from another file:

list<tNode<T>*>::iterator it();  //iterate through the SubNodes

编译器给了我这个错误信息:Tree.h:38:17: error: need 'typename' before 'std::list<tNode<T>*>::iterator' 因为 'std::list*>' 是依赖作用域

The compiler is giving me this error message: Tree.h:38:17: error: need ‘typename’ before ‘std::list<tNode<T>*>::iterator’ because ‘std::list<tNode<T>*>’ is a dependent scope

我不知道为什么编译器会为此对我大喊大叫.

I have no idea why the compiler is yelling at me for this.

推荐答案

在list*>::iterator中，你有一个依赖名称，即依赖于模板参数的名称.

In list<tNode<T>*>::iterator, you have a dependant name, that is, a name that depends on a template parameter.

因此，编译器无法检查 list*>(此时它没有定义)，因此它不知道 >list*>::iterator 要么是静态字段，要么是类型.

As such, the compiler can't inspect list<tNode<T>*> (it doesn't have its definition at this point) and so it doesn't know whether list<tNode<T>*>::iterator is either a static field or a type.

在这种情况下，编译器假定它是一个字段，因此在您的情况下它会产生语法错误.要解决这个问题，只需在声明前放置一个 typename 来告诉编译器它是一个类型:

In such a situation, the compiler assumes that it is a field, so in your case it yields a syntax error. To solve the issue, just tell the compiler that it is a type by putting a typename ahead of the declaration:

typename list<tNode<T>*>::iterator it

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