在模板虚拟类层次结构中调用基类方法 [英] Calling a base class method in a template virtual class hierarchy

问题描述

假设我有以下类层次结构：

Let's say I have the following class hierarchy:

template< class T >
class TestBase {

public:

    virtual T const & do_foo() = 0;

};

template< class T >
class TestDerived : public virtual TestBase< T > {

public:

    virtual int do_bar() {
        return do_foo() + 1;
    }

};

GCC会输出以下内容：

GCC spits out the following:

error: there are no arguments to ‘do_foo’ that depend on a template parameter, so a declaration of ‘do_foo’ must be available [-fpermissive]
note: (if you use ‘-fpermissive’, G++ will accept your code, but allowing the use of an undeclared name is deprecated)

现在，如果我改变这个从一个已知类型（例如 class TestDerived：public virtual TestBase< int> ）实例化的TestBase派生，该代码片段编译正好，所以问题显然与在编译时未知的基类类型有关，但由于我没有实例化任何类的对象，我不明白为什么这应该是重要的。

Now, if I change this to derive from a TestBase instantiated from a known type (e.g. class TestDerived : public virtual TestBase< int >, this snippet compiles just fine so the problem apparently has to do with the base class type being unknown at compile time. But since I haven't instantiated objects of either class anywhere, I don't see why this should even matter.

基本上，我如何解决这个错误，而不诉诸 -fpermissive ？

Basically, how would I resolve the error without resorting to -fpermissive?

推荐答案

解析模板时，现在在实例化模板时，将查找非依赖名称（即不依赖于模板参数的名称）。您需要使 do_foo 一个依赖名称。基本上有两种方法来实现这一点：

Non-dependent names (that is, names which do not depend on template arguments), are looked up when parsing the template, now when instantiating it. You need to make do_foo a dependent name. There are basically two ways to achieve this:

virtual int do_bar() {
    return this->do_foo() + 1;
}

或

template< class T >
class TestDerived : public virtual TestBase< T > {

public:
    using TestBase<T>::do_foo;

    virtual int do_bar() {
        return do_foo() + 1;
    }

};

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