C语言怎么用数组实现反弹球消砖块

本文小编为大家详细介绍"C语言怎么用数组实现反弹球消砖块"，内容详细，步骤清晰，细节处理妥当，希望这篇"C语言怎么用数组实现反弹球消砖块"文章能帮助大家解决疑惑，下面跟着小编的思路慢慢深入，一起来学习新知识吧。

一、效果展示：

二、代码如下：

#include#include#include#include#include#define High 24   //游戏画面尺寸#define Width 36//全局变量int ball_x,ball_y;//小球的坐标int ball_vx,ball_vy;//小球的速度int canvas[High][Width]={0};int position_x,position_y;//挡板的中心坐标int ridus;//挡板的半径大小int left,right;//挡板的左右大小int score=0;//分数//二维数组存储游戏画布中对应的元素//0为空格，1为小球 2为挡板 3为砖块(1分)   4为砖块(2分)  5为砖块(3分)void gotoxy(int x,int y)//将光标移动到(x,y)位置{    HANDLE handle=GetStdHandle(STD_OUTPUT_HANDLE);    COORD pos;    pos.X=x;    pos.Y=y;    SetConsoleCursorPosition(handle,pos);}void startup() //数据的初始化{    int k,i;    ridus=5;    position_x=High-1;    position_y=Width/2;    left=position_y-ridus;    right=position_y+ridus;    ball_x=position_x-1;    ball_y=position_y;    ball_vx=-1;    ball_vy=1;    canvas[ball_x][ball_y]=1;    for(k=left;k<=right;k++)//挡板        canvas[position_x][k]=2;    srand(time(NULL));    for(k=0;k=left)&&(ball_y<=right) )//被挡板挡住        {            printf("\a");//响铃        }        else        {            printf("游戏失败\n");            system("pause");            exit(0);        }    }    if(speed<5)//起到延时的效果        speed++;    if(speed==5)    {        speed=0;        canvas[ball_x][ball_y] = 0;        //更新小球的坐标        ball_x=ball_x+ball_vx;        ball_y=ball_y+ball_vy;        canvas[ball_x][ball_y] = 1;        //碰到边界后反弹        if( (ball_x==0) || (ball_x==High-2) )            ball_vx = -ball_vx;        if( (ball_y==0) || (ball_y==Width-1) )            ball_vy = -ball_vy;        //碰到砖块后反弹        if(canvas[ball_x-1][ball_y]>=3&&canvas[ball_x-1][ball_y]<=5)        {            if(canvas[ball_x-1][ball_y]==3)//判断砖块的类型                score++;            if(canvas[ball_x-1][ball_y]==4)                score=score+2;            if(canvas[ball_x-1][ball_y]==5)                score=score+3;            ball_vx=-ball_vx;            canvas[ball_x-1][ball_y]=0;            printf("\a");        }    }}void updateWithInput()//与用户输入有关的更新{    char input;    if(kbhit())//判断是否有输入    {        input=getch();        if( ((input=='a')||(input=='A')) && (left>0) )        {            canvas[position_x][right]=0;            position_y--;            left=position_y-ridus;            right=position_y+ridus;            canvas[position_x][left]=2;        }        if( ((input=='d')||(input=='D')) && (right
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