C语言中怎么用easyx实现消砖块游戏

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一、最终效果展示

效果图如下：

二、绘制静态的挡板

代码如下:

#include#include#define High 480 //游戏画面尺寸#define Width 640//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 radius;//小球的半径int bar_x,bar_y;//挡板的中心坐标int bar_high,bar_width;//挡板的高度和宽度int bar_left,bar_right,bar_top,bar_bottom;//挡板的左右上下位置坐标void startup()//数据的初始化{    ball_x=Width/2;    ball_y=High/2;    ball_vx=1;    ball_vy=1;    radius=20;    bar_high=High/20;    bar_width=Width/5;    bar_x=Width/2;    bar_y=High-bar_high/2;    bar_left=bar_x-bar_width/2;    bar_right=bar_x+bar_width/2;    bar_top=bar_y-bar_high/2;    bar_bottom=bar_y+bar_high/2;    initgraph(Width,High);    BeginBatchDraw();}void clean()//显示画面{    setcolor(BLACK);//绘制黑线，黑色填充的圆    setfillcolor(BLACK);    fillcircle(ball_x,ball_y,radius);    bar(bar_left,bar_top,bar_right,bar_bottom);//绘制黑色，黑色填充的挡板}void show()//显示画面{    setcolor(YELLOW);//绘制黄线，绿色填充的圆    setfillcolor(GREEN);    fillcircle(ball_x,ball_y,radius);    bar(bar_left,bar_top,bar_right,bar_bottom);//绘制黄色，绿色填充的挡板    FlushBatchDraw();    Sleep(3);}void updateWithoutInput()//与用户输入无关的更新{        ball_x=ball_x+ball_vx;        ball_y=ball_y,ball_vy;//更新小球的坐标        if( (ball_x<=radius)||(ball_x>=Width-radius))            ball_vx=-ball_vx;        if( (ball_y<=radius)||(ball_y>=High-radius))            ball_vy=-ball_vy;}void updateWithInput()//与用户输入有关的更新{}void gameover(){    EndBatchDraw();    closegraph();}int main(){    startup();//数据的初始化    while(1)    {        clean();//把之前绘制的内容清除        updateWithoutInput();//与用户输入无关的更新        updateWithInput();//与用户输入有关的更新        show();//显示新画面    }}

效果图如下:

三、控制挡板

代码如下:

#include#include#define High 480 //游戏画面尺寸#define Width 640//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 radius;//小球的半径int bar_x,bar_y;//挡板的中心坐标int bar_high,bar_width;//挡板的高度和宽度int bar_left,bar_right,bar_top,bar_bottom;//挡板的左右上下位置坐标void startup()//数据的初始化{    ball_x=Width/2;    ball_y=High/2;    ball_vx=1;    ball_vy=1;    radius=20;    bar_high=High/20;    bar_width=Width/5;    bar_x=Width/2;    bar_y=High-bar_high/2;    bar_left=bar_x-bar_width/2;    bar_right=bar_x+bar_width/2;    bar_top=bar_y-bar_high/2;    bar_bottom=bar_y+bar_high/2;    initgraph(Width,High);    BeginBatchDraw();}void clean()//显示画面{    setcolor(BLACK);//绘制黑线，黑色填充的圆    setfillcolor(BLACK);    fillcircle(ball_x,ball_y,radius);    bar(bar_left,bar_top,bar_right,bar_bottom);//绘制黑色，黑色填充的挡板}void show()//显示画面{    setcolor(YELLOW);//绘制黄线，绿色填充的圆    setfillcolor(GREEN);    fillcircle(ball_x,ball_y,radius);    bar(bar_left,bar_top,bar_right,bar_bottom);//绘制黄色，绿色填充的挡板    FlushBatchDraw();    Sleep(3);}void updateWithoutInput()//与用户输入无关的更新{    //挡板和小球碰撞，小球反弹    if(((ball_y+radius>=bar_top)&&(ball_y+radiusbar_top-bar_high/3)))        if((ball_x>=bar_left)&&(ball_x<=bar_right))            ball_vy=-ball_vy;    ball_x=ball_x+ball_vx;    ball_y=ball_y,ball_vy;//更新小球的坐标    if( (ball_x<=radius)||(ball_x>=Width-radius))        ball_vx=-ball_vx;    if( (ball_y<=radius)||(ball_y>=High-radius))        ball_vy=-ball_vy;}void updateWithInput()//与用户输入有关的更新{    char input;    if(kbhit())    {        input=getch();        if(input=='a'&&bar_left>0)        {            bar_x=bar_x-15;//位置左移            bar_left=bar_x-bar_width/2;            bar_right=bar_x+bar_width/2;        }        if(input=='d'&&bar_right0)        {            bar_y=bar_y-15;//位置左移            bar_top=bar_y-bar_high/2;            bar_bottom=bar_y+bar_high/2;        }        if(input=='s'&&bar_bottom
效果图如下:
四、消砖块
代码如下：
#include#include#define High 480 //游戏画面尺寸#define Width 640#define Brick_num 10//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 radius;//小球的半径int bar_x,bar_y;//挡板的中心坐标int bar_high,bar_width;//挡板的高度和宽度int bar_left,bar_right,bar_top,bar_bottom;//挡板的左右上下位置坐标int isBrickExisted[Brick_num];//每个砖块是否存在，1为存在，0为没有了int brick_high,brick_width;//每个砖块的高度和宽度void startup()//数据的初始化{    ball_x=Width/2;    ball_y=High/2;    ball_vx=1;    ball_vy=1;    radius=20;    bar_high=High/20;    bar_width=Width/5;    bar_x=Width/2;    bar_y=High-bar_high/2;    bar_left=bar_x-bar_width/2;    bar_right=bar_x+bar_width/2;    bar_top=bar_y-bar_high/2;    bar_bottom=bar_y+bar_high/2;    brick_width=Width/Brick_num;    brick_high=High/Brick_num;    int i;    for(i=0;i=bar_top)&&(ball_y+radiusbar_top-bar_high/3)))        if((ball_x>=bar_left)&&(ball_x<=bar_right))            ball_vy=-ball_vy;    ball_x=ball_x+ball_vx;    ball_y=ball_y+ball_vy;//更新小球的坐标    //小球和边界碰撞    if( (ball_x<=radius)||(ball_x>=Width-radius))        ball_vx=-ball_vx;    if( (ball_y<=radius)||(ball_y>=High-radius))        ball_vy=-ball_vy;    //判断小球是否和某个砖块碰撞    int i,brick_left,brick_right,brick_top,brick_bottom;    for(i=0;i=brick_left)&&(ball_x<=                brick_right))            {                isBrickExisted[i]=0;                ball_vy=-ball_vy;            }        }    }}void updateWithInput()//与用户输入有关的更新{    char input;    if(kbhit())    {        input=getch();        if(input=='a'&&bar_left>0)        {            bar_x=bar_x-15;//位置左移            bar_left=bar_x-bar_width/2;            bar_right=bar_x+bar_width/2;        }        if(input=='d'&&bar_right
效果图如下:
五、鼠标交互
先看一个关于鼠标交互的实例
#include#includeint main(void){    initgraph(640,480);//初始化图形窗口    MOUSEMSG m;//定义鼠标消息    while(1)    {        m=GetMouseMsg();//获取一条鼠标消息        if(m.uMsg==WM_MOUSEMOVE)        {            putpixel(m.x,m.y,WHITE);//鼠标移动的时候画小白点        }        else if(m.uMsg==WM_LBUTTONDOWN)        {            rectangle(m.x-5,m.y-5,m.x+5,m.y+5);            //鼠标左键按下时在鼠标位置画一个方块        }        else if(m.uMsg==WM_RBUTTONUP)        {            circle(m.x,m.y,10);            //鼠标右键按下时在鼠标位置画一个圆        }    }    return 0;}
用鼠标控制挡板移动，按鼠标左键初始化小球位置
代码如下:
#include#include#define High 480 //游戏画面尺寸#define Width 640#define Brick_num 10//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 radius;//小球的半径int bar_x,bar_y;//挡板的中心坐标int bar_high,bar_width;//挡板的高度和宽度int bar_left,bar_right,bar_top,bar_bottom;//挡板的左右上下位置坐标int isBrickExisted[Brick_num];//每个砖块是否存在，1为存在，0为没有了int brick_high,brick_width;//每个砖块的高度和宽度void startup()//数据的初始化{    ball_x=Width/2;    ball_y=High/2;    ball_vx=1;    ball_vy=1;    radius=20;    bar_high=High/20;    bar_width=Width/5;    bar_x=Width/2;    bar_y=High-bar_high/2;    bar_left=bar_x-bar_width/2;    bar_right=bar_x+bar_width/2;    bar_top=bar_y-bar_high/2;    bar_bottom=bar_y+bar_high/2;    brick_width=Width/Brick_num;    brick_high=High/Brick_num;    int i;    for(i=0;i=bar_top)&&(ball_y+radiusbar_top-bar_high/3)))        if((ball_x>=bar_left)&&(ball_x<=bar_right))            ball_vy=-ball_vy;    ball_x=ball_x+ball_vx;    ball_y=ball_y+ball_vy;//更新小球的坐标    //小球和边界碰撞    if( (ball_x<=radius)||(ball_x>=Width-radius))        ball_vx=-ball_vx;    if( (ball_y<=radius)||(ball_y>=High-radius))        ball_vy=-ball_vy;    //判断小球是否和某个砖块碰撞    int i,brick_left,brick_right,brick_top,brick_bottom;    for(i=0;i=brick_left)&&(ball_x<=                brick_right))            {                isBrickExisted[i]=0;                ball_vy=-ball_vy;            }        }    }}void updateWithInput()//与用户输入有关的更新{    /*char input;    if(kbhit())    {    input=getch();    if(input=='a'&&bar_left>0)    {    bar_x=bar_x-15;//位置左移    bar_left=bar_x-bar_width/2;    bar_right=bar_x+bar_width/2;    }    if(input=='d'&&bar_right
效果图如下：
感谢各位的阅读，以上就是"C语言中怎么用easyx实现消砖块游戏"的内容了，经过本文的学习后，相信大家对C语言中怎么用easyx实现消砖块游戏这一问题有了更深刻的体会，具体使用情况还需要大家实践验证。这里是，小编将为大家推送更多相关知识点的文章，欢迎关注！