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C语言简易版flappy bird小游戏

2020-05-23 13:23:12
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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:

C语言,flappy bird,小游戏

首先,先画出整个小游戏实现的流程图,如下:

C语言,flappy bird,小游戏

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

 

#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

#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);}

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持VEVB武林网。


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