模板和嵌套类/结构 [英] Templates and nested classes/structures

问题描述

我有一个简单的容器：

template <class nodeType> list {
    public:
        struct node {
            nodeType info;
            node* next;
        };

    //...
};

现在，有一个函数 _search 它搜索列表并返回对匹配的节点的引用。现在，当我指的是函数的返回类型时，我认为应该是 list< nodeType> :: node * 。这是正确的吗？当我定义函数inline时，它工作完美：

Now, there is a function called _search which searches the list and returns a reference to the node which matched. Now, when I am referring to the return-type of the function, I think it should be list<nodeType>::node*. Is this right? When I define the function inline, it works perfectly:

template <class nodeType> list {
    public:
        struct node {
            nodeType info;
            node* next;
        };

        node* _search {
            node* temp;
            // search for the node
            return temp;
        }
};

但是，如果我定义类外的函数，

But, if I define the function outside the class,

template <class nodeType> list<nodeType>::node* list<nodeType>::_search() {
    //function
}

它不工作。编译器给出一个错误，在列表< nodeType> :: _ search 或某事之前说出 Expected构造函数。错误是类似的。我现在没有机器可以测试它。

it doesn't work. The compiler gives an error saying Expected constructor before list<nodeType>::_search or something. The error is something similar to this. I don't have a machine on which I can test it currently.

任何帮助真的很感激。

推荐答案

这是因为节点是一个依赖类型。您需要写如下的签名（注意，为了清楚起见我已将它分成两行）

that's because node is a dependent type. You need to write the signature as follows (note that I have broken it into 2 lines for clarity)

template <class nodeType> 
typename list<nodeType>::node* list<nodeType>::_search() 
{
    //function
}

请注意使用 typename 关键字。

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