具有相同名称的typedef和template参数 [英] typedef and template parameter with same name
问题描述
为什么这种情况不正确(这是合乎逻辑的)
Why is that case incorrect (it's logical)
template <typename T>
struct Der: public Base
{
typedef int T;
T val;
};
,但是这种情况正确吗?
, but that case is correct?
struct Base
{
typedef int T;
};
template <typename T>
struct Der: public Base
{
T val;
};
标准14.6.1/7说:
The Standard 14.6.1/7 says:
对于不依赖于模板参数(14.6.2)的每个基类,如果类模板的定义或出现在模板定义之外的此类模板的成员的定义,如果基类的名称或基类成员的名称与模板参数的名称相同,基类名称或成员名称隐藏了模板参数的名称(3.3.7).
In the definition of a class template or in the definition of a member of such a template that appears outside of the template definition, for each base class which does not depend on a template-parameter (14.6.2), if the name of the base class or the name of a member of the base class is the same as the name of a template-parameter, the base class name or member name hides the template-parameter name (3.3.7).
为什么这里没有歧义?
推荐答案
根据[temp.local]/6,第一个示例不正确:
The first example is incorrect according to [temp.local]/6:
不得在其范围(包括嵌套范围)内重新声明 template-parameter .
但是,在
template <typename T>
struct Der: public Base
{
T val;
};
T
被从 Base
继承的名称隐藏-如您的报价 所指定.
T
is hidden by the name inherited from Base
- as specified by your quote.
[..] 如果基类的名称或成员的名称基类与 template-parameter 的名称相同,基类名称或成员名称隐藏 template-parameter 名称(3.3.7).
[..] if the name of the base class or the name of a member of the base class is the same as the name of a template-parameter, the base class name or member name hides the template-parameter name (3.3.7).
也就是说,成员 val
的类型为 int
. 演示 .
That is, the member val
is of type int
. Demo.
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