#include <cstdio>

int main ()
 {
   char *near_title = "Jamsa's 1001 C & C++ Tips";
   char far *far_title = "Jamsa's 1001 C & C++ Tips";

   printf("The book's title: %Ns\n", near_title);
   printf("The book's title: %Fs\n", far_title);
   return 0;
 }
运行时出现了三个错误，搜了好久，不知道应该怎么做才好，希望大家基于一点帮助
1>------ 已启动生成: 项目: lmy_console, 配置: Debug Win32 ------
1>  20110407.cpp
1>d:\users\yaoyao\documents\visual studio 2010\projects\lmy_win32\lmy_console\lmy_console\20110407.cpp(6): error C2143: 语法错误 : 缺少“;”(在“*”的前面)
1>d:\users\yaoyao\documents\visual studio 2010\projects\lmy_win32\lmy_console\lmy_console\20110407.cpp(6): error C2065: “far_title”: 未声明的标识符
1>d:\users\yaoyao\documents\visual studio 2010\projects\lmy_win32\lmy_console\lmy_console\20110407.cpp(9): error C2065: “far_title”: 未声明的标识符
========== 生成: 成功 0 个，失败 1 个，最新 0 个，跳过 0 个 ==========

#include <cstdio>

int main ()
{
   char *near_title = "Jamsa's 1001 C & C++ Tips";
   char  *far_title = "Jamsa's 1001 C & C++ Tips";//你多写了个far
   printf("The book's title: %Ns\n", near_title);
   printf("The book's title: %Fs\n", far_title);
   return 0;
}

#include <cstdio>
int main ()
{
   char *near_title = "Jamsa's 1001 C & C++ Tips";
   char *far_title = "Jamsa's 1001 C & C++ Tips";  //无指针，没听过!

   printf("The book's title: %Ns\n", near_title);
   printf("The book's title: %Fs\n", far_title);
   return 0;
}

保护模式下似乎没有远近指针之分。

什么叫做远近指针，请各路好汉明示下