C语言简易版小游戏【flappy bird】

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 
#include 
#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);
}