假期在家無聊，想隨便碼點東西，故有此簡陋的小游戲誕生。覺著可能對初學C語言的小伙伴練習有點幫助，故寫此博客。游戲界面如下：
首先，先畫出整個小游戲實現的流程圖，如下：
思路很簡單，整個游戲界面是由一個大的char類型數組構成，更新數組的值然后不停的打印出來就形成了動態效果。
由上圖看，大循環是保證游戲一直不斷的進行下去，小循環是讓小鳥的速度大于游戲界面里背景（由#構成的柱子）的速度（小鳥動四下柱子才動一下）。
下面是具體代碼（水平有限大家多多見諒，但是效果還是有的！）
Bird.c文件
#include <stdio.h>#include <windows.h>#include "Interface.h"int main(void){ InitialInterface(); for(;;) { newinterface(); scoring();//過一個柱子計一次分，所以和柱子更新速度一致 for (int i = 0; i < 4; i++)//小鳥的速度是柱子的4倍 { birdmove(); draw(); Sleep(50); } } return 0;}Interface.h文件
#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
#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 }, };//初始界面矩陣，ASCII碼中“ ”是32，“&”是38表示小鳥，“#”是35用來畫柱子int num = 0;//用于計數輸出并排兩列黑柱子同一位置int black;//黑方塊位置int p= M/2 ;//小鳥初始位置int score = 0;//分數/*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/"使小鳥跳起來，別落地，順利穿過盡可能多的柱子!/n"); for (int i = 0; i < M; i++) { printf(" "); for (int j = 0; j < N; j++) { printf("%c", interf[i][j]); } printf("/n"); }}/*更新界面各個柱子*/void newinterface(void){ if (interf[0][1] == 35 && num==0)//當矩陣第二列為黑色方塊時，計算出下一次黑柱子上半部分的位置 { black = 5 + rand() % 5; num = 2;//黑柱子是兩列#組成，第二列與第一列位置一樣，用num保證兩列位置一致 } 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)))//此時上面的第二列變成了第一列，更新下一個黑柱子，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--;}/*更新小鳥位置*/void birdmove(void){ for (int a = 0; a < 3; a++) { if (a == 2 && p > 0)//減緩鳥的速度，使按鍵上跳速度是下落的4倍 { p = p + 1; } if (_kbhit()) { if (_getch() == 'w' || _getch() == 'W') { p = p - 3; } } }}/*計分*/void scoring(void){ if (p > 20 || interf[p][0] == 35) { system("cls"); printf("/n/n 游戲結束!/n/n"); printf(" 最終得分：%d/n/n/n", score); system("pause"); } if (interf[0][0] == 35 && interf[0][1] == 32 ) score++;}/*重畫界面*/void draw(void){ system("cls"); printf("/n 作者：xhyang，博客地址：http://blog.csdn.net/weixin_39449570/n"); printf(" 按/"w/"使小鳥跳起來，別落地，順利穿過盡可能多的柱子!/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);}以上就是本文的全部內容，希望對大家的學習有所幫助，也希望大家多多支持VEVB武林網。
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