模板函数返回模板类型

Question

我有以下功能,但我的编译器(VS2003)说,分配T = ....是非法的.有人可以澄清我做错了什么吗？值的类型是boost :: variant.node是一个结构.

 template <typename T>
    T find_attribute(const std::string& attribute)
    {

        std::vector<boost::shared_ptr<node> >::iterator nodes_iter = _request->begin();

        for (; nodes_iter != _request->end(); nodes_iter++)
        {
            std::vector<node::attrib>::iterator att_iter = (*nodes_iter)->attributes.begin();
            for (; att_iter != att_iter; (*nodes_iter)->attributes.end())
            {
                if ((*att_iter).key.compare(attribute) == 0) {
                    T = (*att_iter).value;  //T : Illegal use of this type as an expression.
                                    return T; 
                            }

            }

        }
    }

Answer 1

你应该声明一个变量:

if ((*att_iter).key.compare(attribute) == 0) {
    T temp = (*att_iter).value;  //T : Illegal use of this type as an expression.
    return temp; 
}