假期在家无聊，想随便码点东西，故有此简陋的小游戏诞生。觉着可能对初学C语言的小伙伴练习有点帮助，故写此博客。游戏界面如下：
首先，先画出整个小游戏实现的流程图，如下：
思路很简单，整个游戏界面是由一个大的char类型数组构成，更新数组的值然后不停的打印出来就形成了动态效果。
由上图看，大循环是保证游戏一直不断的进行下去，小循环是让小鸟的速度大于游戏界面里背景（由#构成的柱子）的速度（小鸟动四下柱子才动一下）。
下面是具体代码（水平有限大家多多见谅，但是效果还是有的！）
Bird.c文件
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#include <stdio.h>#include <windows.h>#include "Interface.h"int main(void){ InitialInterface(); for(;;) { newinterface(); scoring();//过一个柱子计一次分，所以和柱子更新速度一致 for (int i = 0; i < 4; i++)//小鸟的速度是柱子的4倍 { birdmove(); draw(); Sleep(50); } } return 0;} |
Interface.h文件
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#ifndef INTERFACE_H#define INTERFACE_H#define M 20#define N 36void InitialInterface(void);void newinterface(void);void birdmove(void);void scoring(void);void draw(void);#endif |
Interface.c文件
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#include <stdio.h>#include <stdlib.h>#include<conio.h>#include "interface.h"char interf[M][N] = {{ 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,32,32 }, { 32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,32,32 }, { 38,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32 }, { 32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, { 32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35,32,32,32,32,32,32,32,32,32,32,35,35 }, };//初始界面矩阵，ASCII码中“ ”是32，“&”是38表示小鸟，“#”是35用来画柱子int num = 0;//用于计数输出并排两列黑柱子同一位置int black;//黑方块位置int p= M/2 ;//小鸟初始位置int score = 0;//分数/*初始化界面*/void InitialInterface(void){ printf("\n 作者：xhyang，博客地址：http://blog.csdn.net/weixin_39449570\n"); printf(" 按\"w\"使小鸟跳起来，别落地，顺利穿过尽可能多的柱子!\n"); for (int i = 0; i < M; i++) { printf(" "); for (int j = 0; j < N; j++) { printf("%c", interf[i][j]); } printf("\n"); }}/*更新界面各个柱子*/void newinterface(void){ if (interf[0][1] == 35 && num==0)//当矩阵第二列为黑色方块时，计算出下一次黑柱子上半部分的位置 { black = 5 + rand() % 5; num = 2;//黑柱子是两列#组成，第二列与第一列位置一样，用num保证两列位置一致 } for (int i = 0; i < M; i++) { for (int j = 0; j < N - 1; j++) { interf[i][j] = interf[i][j + 1]; } if (interf[0][0] == 35 && (i < black || i>(black + 5)))//此时上面的第二列变成了第一列，更新下一个黑柱子，有了黑柱子上半部分位置+5即是下半部分的起始位置 { interf[i][N-1] = 35; } else { interf[i][N-1] = 32; } } if (num > 0) num--;}/*更新小鸟位置*/void birdmove(void){ for (int a = 0; a < 3; a++) { if (a == 2 && p > 0)//减缓鸟的速度，使按键上跳速度是下落的4倍 { p = p + 1; } if (_kbhit()) { if (_getch() == 'w' || _getch() == 'W') { p = p - 3; } } }}/*计分*/void scoring(void){ if (p > 20 || interf[p][0] == 35) { system("cls"); printf("\n\n 游戏结束!\n\n"); printf(" 最终得分：%d\n\n\n", score); system("pause"); } if (interf[0][0] == 35 && interf[0][1] == 32 ) score++;}/*重画界面*/void draw(void){ system("cls"); printf("\n 作者：xhyang，博客地址：http://blog.csdn.net/weixin_39449570\n"); printf(" 按\"w\"使小鸟跳起来，别落地，顺利穿过尽可能多的柱子!\n"); for (int i = 0; i < M; i++) { printf(" "); for (int j = 0; j < N; j++) { if (i == p && j == 0 && interf[p][0] != 35) printf("%c", 38); else printf("%c", interf[i][j]); } printf("\n"); } printf(" 得分：%d \n", score);} |
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原文链接：https://blog.csdn.net/weixin_39449570/article/details/79307447
