模板内部类型定义 typename::int

Designating Types inside the Template Definition

在模板定义内部指定类型

In addition to defining data or function members, a class may define type members. For example, the library container classes define various types, such as size_type, that allow us to use the containers in a machine-independent way. When we want to use such types inside a function template, we must tell the compiler that the name we are using refers to a type. We must be explicit because the compiler (and a reader of our program) cannot tell by inspection when a name defined by a type parameter is a type or a value. As an example, consider the following function:

除了定义数据成员或函数成员之外，类还可以定义类型成员。例如，标准库的容器类定义了不同的类型，如 size_type，使我们能够以独立于机器的方式使用容器。如果要在函数模板内部使用这样的类型，必须告诉编译器我们正在使用的名字指的是一个类型。必须显式地这样做，因为编译器（以及程序的读者）不能通过检查得知，由类型形参定义的名字何时是一个类型何时是一个值。例如，考虑下面的函数：

     template 
     Parm fcn(Parm* array, U value)
     {
         Parm::size_type * p; // If Parm::size_type is a type, then a declaration
                              // If Parm::size_type is an object, then multiplication
     }

 

We know that size_type must be a member of the type bound to Parm, but we do not know whether size_type is the name of a type or a data member. By default, the compiler assumes that such names name data members, not types.

我们知道 size_type 必定是绑定到 Parm 的那个类型的成员，但我们不知道 size_type 是一个类型成员的名字还是一个数据成员的名字，默认情况下，编译器假定这样的名字指定数据成员，而不是类型。

 

If we want the compiler to treat size_type as a type, then we must explicitly tell the compiler to do so:

如果希望编译器将 size_type 当作类型，则必须显式告诉编译器这样做：

     template 
     Parm fcn(Parm* array, U value)
     {
         typename Parm::size_type * p; // ok: declares p to be a pointer
     }

 

We tell the compiler to treat a member as a type by prefixing uses of the member name with the keyword typename. By writing typename Parm::size_type we say that member size_type of the type bound to Parm is the name of a type. Of course, this declaration puts an obligation on the types used to instantiate fcn: Those types must have a member named size_type that is a type.

通过在成员名前加上关键字 typename 作为前缀，可以告诉编译器将成员当作类型。通过编写 typename parm::size_type，指出绑定到 Parm 的类型的 size_type 成员是类型的名字。当然，这一声明给用实例化 fcn 的类型增加了一个职责：那些类型必须具有名为 size_type 的成员，而且该成员是一个类型。

If there is any doubt as to whether typename is necessary to indicate that a name is a type, it is a good idea to specify it. There is no harm in specifying typename before a type, so if the typename was unnecessary, it won't matter. 如果拿不准是否需要以 typename 指明一个名字是一个类型，那么指定它是个好主意。在类型之前指定 typename 没有害处，因此，即使 typename 是不必要的，也没有关系。

 

---------c++ primer