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，原版试卷既为原型；
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; 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) { paper_t origin_paper = {"Li", 5, "1 + 1 = ?", paper_copy}; for(int i = 1; i <= 10; i++) { paper_t* VAR(paper, i) = clone(&origin_paper); printf("paper_%d => name: %s, class: %d, question: %srn", 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 = ?
