C + + Template programming-fifth Chapter technical Basics

Keyword Typename

1 template<typename t>2class  someclass3{4     TypeName T::subtype * ptr; 5 };

If no typename,t::subtype is considered to be a static member.

A Practical Example:

1 //print elements in a STL container2Template<typename t>3 voidPrint (TConst&con)4 {5 TypeName T::const_iterator Pos;6 TypeName T::const_iterator End (Con.end ());7     8      for(Pos=con.begin (); pos!=end; pos++)9         ....Ten}

If you are in a template and want to invoke another template function, you may need to use the. Template

1 template<int n>2voidconst & BS)3{  4     std::cout<<bs.template to_string<char, char_traits<Char allocator<char> > ();  // Here we are inside a template and calling a template function 5 }

In verse 5.2, the book's statement and my running results diverge, my operating environment is vs2013.

1#include <iostream>2 3 using namespacestd;4 5 voidfoo ()6 {7cout <<"Global Foo"<<Endl;8 }9 TenTemplate<typename t> One classbase{ A  Public: -     voidfoo () -     { thecout <<"Foo in Base"<<Endl; -     } - }; -  +Template<typename t> - classDerived1:base < T > + { A  Public: at     voidgoo () -     { -Foo ();//Foo in Base -Base<t>::foo ();//Foo in Base -:: foo ();//Global Foo -     } in }; -  to intMain () + { -derived1<int>D1; the D1.goo (); *     return 0; $}

The book says that the first Foo call in Goo is not actually called Foo in base, but my running results negate the book's writing. So it might be a matter of compiling the environment. To avoid uncertainty, add base<t>:: prefix, or use this pointer to compare insurance, to avoid uncertainty.

Changlog:

2015/6/18 to scroll to see the mathematical equation, first write so much.

C + + Template programming-fifth Chapter technical Basics