关于c ++：嵌套类是类模板中的依赖类型？


Nested classes are dependent types in class templates?

考虑下面的：P＞＜／

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template<class> struct T
{
    struct T1
    {
        struct T2
        {

        };
    };
    /*typename*/ T1::T2 m;
};

没有typename磁带T2是失败的事情考虑，在AS和依赖thus not a name，type。后看到C + +标准草案(17，n4659)矛盾：为什么我相信我P＞＜／

§17.6.2.1？5P＞＜／

A name is a member of the current instantiation if it is
— An unqualified name that, when looked up, refers to at least one member of a class that is the current instantiation or a non-dependent base class thereof.
...
A name is a dependent member of the current instantiation if it is a member of the current instantiation that, when looked up, refers to at least one member of a class that is the current instantiation.

T1是依赖人of the current实例化。T2 is not of the current成员实例化(EN是T1，学院成员)P＞＜／

§17.6.2.1？9.3P＞＜／

A type is dependent if it is
...
— a nested class or enumeration that is a dependent member of the current instantiation,
...

T1是在套式舱和thus依赖型。P＞＜／

§17.6？3P＞＜／

When a qualified-id is intended to refer to a type that is not a member of the current instantiation (17.6.2.1) and its nested-name-specifier refers to a dependent type, it shall be prefixed by the keyword typename, forming a typename-specifier. ...

thus，typenameis needed。P＞＜／

我的理解是正确的吗？如果我知道这背后的理由，什么茶？我永远T1::T2of how the other查找find anything在T2比T1套式？P＞＜／

是的，你是对的。

在您的例子中，这并不重要，因为您不可能在不实例化EDOCX1(因为它是类成员)的情况下专门化T1。但你可以这样做，把m放在一个函数中：

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template<class> struct T
{
    struct T1
    {
        struct T2
        {

        };
    };

    void foo() {
        typename T1::T2 m;
    }
};

template<>
struct T<void>::T1 {
    int T2;
};

如果T1不依赖，您的代码可能会改变含义，因为T2引用的是值而不是类型。