C语言简易版flappy bird小游戏
时间：2020-12-22 18:21:47|栏目：C代码|点击： 次
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
#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);
}
