C语言怎么实现简易版flappy bird小游戏

小编给大家分享一下C语言怎么实现简易版flappy bird小游戏，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！

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

#include<stdio.h>
#include<windows.h>
#include"Interface.h"

intmain(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(inti=0;i<4;i++)//小鸟的速度是柱子的4倍
{
birdmove();
draw();
Sleep(50);
}
}
return0;
}

Interface.h文件

#ifndefINTERFACE_H
#defineINTERFACE_H

#defineM20
#defineN36

voidInitialInterface(void);
voidnewinterface(void);
voidbirdmove(void);
voidscoring(void);
voiddraw(void);

#endif

Interface.c文件

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include"interface.h"


charinterf[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

intnum=0;//用于计数输出并排两列黑柱子同一位置
intblack;//黑方块位置
intp=M/2;//小鸟初始位置
intscore=0;//分数

/*初始化界面*/
voidInitialInterface(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(inti=0;i<M;i++)
{
printf("");
for(intj=0;j<N;j++)
{
printf("%c",interf[i][j]);
}
printf("\n");
}
}


/*更新界面各个柱子*/
voidnewinterface(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(inti=0;i<M;i++)
{

for(intj=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*/
voidbirdmove(void)
{
for(inta=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*/
voidscoring(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*/
voiddraw(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(inti=0;i<M;i++)
{
printf("");
for(intj=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);
}

以上是“C语言怎么实现简易版flappy bird小游戏”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注恰卡编程网行业资讯频道！