设计模式之原型模式

　　介绍：
　　原型模式是一种创建型设计模式，使你可以通过复制原型来创建一个一模一样的新的对象，无需考虑创建过程，数据类型等内容；
　　作用：
　　创建一个与原型一模一样的对象；
　　解决问题： 即使原型的状态有所改变，也可以创建出与其当前状态一致的对象  省去了创建对象的复杂操作，不必关注创建对象的具体过程
　　类比：
　　老师打印试卷，拿着原版试卷，可以复制出任意数量的试卷，原版试卷即为原型；
　　代码示例： #include  #include  #include   #define VAR(name, id)   name##id  typedef struct Paper {     char* teacher_name;     int class;     char* question;     struct Paper* (*copy)(struct Paper* pPaper); } paper_t;   /*原型的复制接口*/ paper_t* paper_copy(paper_t* pPaper) {     paper_t* copy = (paper_t*)malloc(sizeof(paper_t));     memcpy(copy, pPaper, sizeof(paper_t));     return copy; }   paper_t* clone(paper_t* pPaper) {     return pPaper->copy(pPaper); }  void main(void) {     /*定义原型*/     paper_t origin_paper = {＂Li＂, 5, ＂1 + 1 = ?＂, paper_copy};          /*复制10份试卷*/     for(int i = 1; i <= 10; i++)     {         paper_t* VAR(paper, i) = clone(&origin_paper);         printf(＂paper_%d => name: %s, class: %d, question: %sr ＂,                 i, VAR(paper, i)->teacher_name, VAR(paper, i)->class, VAR(paper, i)->question);     } }
　　输出： paper_1 => name: Li, class: 5, question: 1 + 1 = ? paper_2 => name: Li, class: 5, question: 1 + 1 = ? paper_3 => name: Li, class: 5, question: 1 + 1 = ? paper_4 => name: Li, class: 5, question: 1 + 1 = ? paper_5 => name: Li, class: 5, question: 1 + 1 = ? paper_6 => name: Li, class: 5, question: 1 + 1 = ? paper_7 => name: Li, class: 5, question: 1 + 1 = ? paper_8 => name: Li, class: 5, question: 1 + 1 = ? paper_9 => name: Li, class: 5, question: 1 + 1 = ? paper_10 => name: Li, class: 5, question: 1 + 1 = ?