问题描述

这是重现我的问题的代码示例:

Here is code sample which reproduces my problem:

template <typename myType>
class Base {
public:
    Base() {}
    virtual ~Base() {}
protected:
    int myOption;
    virtual void set() = 0;
};

template <typename InterfaceType>
class ChildClass : public Base < std::vector<InterfaceType> >
{
public:
    ChildClass() {}
    virtual ~ChildClass() {}
 protected:
    virtual void set();
};

template <typename InterfaceType>
void ChildClass<InterfaceType>::set()
{
     myOption = 10;
}

我在main()中的用法:

ChildClass<int> myObject;

我收到以下错误(ubuntu 上的 gcc 4.4.3):

I get the following error (gcc 4.4.3 on ubuntu):

'myOption' 未在此范围内声明

‘myOption’ was not declared in this scope

如果我的 ChildClass 没有新的模板参数，这可以正常工作，即:

If my ChildClass would be without new template parameter this would work fine, i.e.:

class ChildClass : public Base < std::vector<SomeConcreteType> >

<小时>

编辑

我已经设法解决它，如果我的 set 方法看起来像:

Edit

I've managed to solve it, if my set method looks like:

Base<std::vector<InterfaceType> >::myOption = 10;

它工作正常.仍然不确定为什么我需要指定所有模板参数.

It works fine. Still though not sure why I need to specify all template parameters.

推荐答案

myOption 不是依赖名称，即它不显式依赖模板参数，因此编译器会尝试查找它早期的.您必须将其设为依赖名称:

myOption is not a dependent name, i.e. it doesn't depend on the template arguments explicitly so the compiler tries to look it up early. You must make it a dependent name:

template <typename InterfaceType>
void ChildClass<InterfaceType>::set()
{
     this->myOption = 10;
}

现在它取决于 this 的类型，因此取决于模板参数.因此编译器会在实例化时绑定它.

Now it depends on the type of this and thus on the template arguments. Therefore the compiler will bind it at the time of instantiation.

这称为两阶段名称查找.

这篇关于“未在此范围内声明"模板和继承错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！