Effective C + + clause 42

In this section we discuss the difference between the class keyword and the TypeName keyword and the effect on template functions.

For example, the following code:

T>T getValue1(T m){    return2T>T getValue2(T m){    return2;}

In the code above. The function of class and TypeName is the same. No matter what the difference. So the program ape is able to use class or typename when declaring templates according to personal habits.

It is only in some cases that you can use TypeName instead of class.

For example, the following code:

 #include <iostream>  using  namespace  STD ; class  Defalut{public : typedef  int  A;}; template  <typename  t> int  getValue (T m) {typename  t::a a1;//assumes that there is no TypeName compilation error  a1 = 0 ; return  A1;}    int  Main () {Defalut D;    cout  <<getvalue (d); return  0 ;}  

C + + stipulates that the gaze line above can only add typename, not add Class
Why not add the TypeName program will go wrong, we now imagine, for example, the following control code form:

classDefalut{public:    typedef int a;};classDefalut{public:    static int a;};Defalut::a;

Now let's look at the call of the two classes with the same name, a return type int, and a return integer value, defalut::a.
So that means that the t::a inference is the return value or type in the template. For C + +. Without the predecessor Keyword TypeName think it is the return value. such as T::a, feel the return of a static value.

Added the TypeName keyword to think of as a return type, such as TypeName T::a. Think of returning a nested type.
Nested types are types defined in class classes or other structures.

Effective C + + clause 42