views:

96

answers:

2

I have a simple container :

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

    //...
};

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;
        }
};

But, if I define the function outside the class,

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

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.

Any help is sincerely appreciated.

+9  A: 

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
}

Note the use of the typename keyword.

rlbond
More, very geeky, details can be found in the C++ FAQ Lite: http://www.parashift.com/c++-faq-lite/templates.html#faq-35.18
Michael Dunn
thanks a lot for the help.. it works perfectly now. thanks again..
Rohan Prabhu
+2  A: 

You need to tell the compiler that node is a type using the keyword typename.Otherwise it will think node as a static variable in class list. Add typename whenever you use node as a type in your implementation of list.

Ponting