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"

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 36

void 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);
}

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