本篇内容主要讲解“C语言基于EasyX如何实现贪吃蛇游戏”，感兴趣的朋友不妨来看看。本文介绍的方法操作简单快捷，实用性强。下面就让小编来带大家学习“C语言基于EasyX如何实现贪吃蛇游戏”吧!
成品展示：
实现思路：
贪吃蛇的实现思路并不复杂，由于我们需要将数据展示在图形窗口上，因此就不需要之前那种用数组表示整个游戏地图的方法。
贪吃蛇的蛇有X坐标和Y坐标，而且不止一节，因此需要一个坐标结构体数组来保存，蛇的移动思路是除了第一节以外，后面每一节都是前面一节的坐标，然后通过键盘的输入输出使蛇的X或Y坐标加或者减来起到上下左右移动的作用。通过判断蛇与食物的坐标是否重合绝对是否吃到食物，吃到食物以后，蛇的长度加一，分数增加，食物重新生成。通过与自身坐标比较和界面坐标比较，判断是否撞墙或者撞自己，此时游戏结束。一直循环移动并且在图形界面展示，其中通过睡眠函数Sleep来控制打印速度，从而控制蛇的移动速度。
最后如果需要在游戏时开启音乐，为了方便则需要在文件目录下放入音乐文件。
代码实现
game.h
#pragma once#include<stdio.h>#include<time.h>#include<stdlib.h>#include<graphics.h>#include<conio.h>#include<windows.h>#include<mmsystem.h>#pragma comment(lib,"Winmm.lib")#define length 100 //蛇长度//1.数据分析：基本数据成员分析//坐标属性，蛇属性，食物属性//坐标属性typedef struct pointXY{ int x; int y;}MYPOINT;//蛇属性struct Snake{ int num;//蛇的节数 MYPOINT xy[length]; //蛇最多有100节 char postion;//蛇的方向};//食物的属性struct Food{ MYPOINT fdxy; int eatgrade=0;//吃食物的分数 int flag=0;//食物是否存在，0表示存在，1表示不存在};//方向enum movePostion{right=77,left=75,down=80,up=72};//主窗口static HWND hwnd = NULL;//2.过程分析：函数分析//初始化蛇和食物void initSnakeFood(struct Snake*snake, struct Food* food);//画蛇void drawSnake(struct Snake* snake);//移动蛇void moveSnake(struct Snake* snake);//按键处理void keyDown(struct Snake* snake);//初始化食物void initFood(struct Food*food, struct Snake* snake);//画食物void drawFood(struct Food* food);//吃食物void eatFood(struct Food* food, struct Snake* snake);//判断是否游戏结束int snakeDie(struct Snake* snake);//判断游戏是否胜利int winGame(struct Snake* snake);//展示分数void showGrade(struct Food* food);//按空格键暂停void pauseMoment();这个头文件主要包含函数声明和头文件的引用。
game.cpp
#undef UNICODE#undef _UNICODE#include"game.h"void initSnakeFood(struct Snake* snake, struct Food* food){ //把前三节的属性写出来，初始化数据 snake->xy[2].x = 0; snake->xy[2].y = 0; snake->xy[1].x = 10; snake->xy[1].y = 0; snake->xy[0].x = 20; snake->xy[0].y = 0; snake->num = 3;//开始有三节 snake->postion = right;//初始化方向为右 food->eatgrade = 0;}void drawSnake(struct Snake* snake){ for (int i = 0; i < snake->num; i++) { setlinecolor(BLACK);//矩形边框线的颜色 setfillcolor(RGB(rand() % 255, rand() % 255, rand() % 255));//内部填充颜色 fillrectangle(snake->xy[i].x, snake->xy[i].y, snake->xy[i].x + 10, snake->xy[i].y + 10); }}void moveSnake(struct Snake* snake){ //除了第一节之外，后面每一节都是前面一节的坐标 for (int i = snake->num; i > 0; i--) { snake->xy[i].x = snake->xy[i - 1].x; snake->xy[i].y = snake->xy[i - 1].y; } //第一节的处理 switch(snake->postion) { case right: snake->xy[0].x += 10; break; case left: snake->xy[0].x -= 10; break; case down: snake->xy[0].y += 10; break; case up: snake->xy[0].y -= 10; default: break; }}void keyDown(struct Snake* snake){ char userkey = 0; userkey = _getch(); switch (userkey) { case right: if (snake->postion != left) snake->postion = right; break; case left: if (snake->postion != right) snake->postion = left; break; case down: if (snake->postion != up) snake->postion = down; break; case up: if (snake->postion != down) snake->postion = up; break; }}void initFood(struct Food* food, struct Snake* snake){ food->fdxy.x = rand() % 64 * 10; food->fdxy.y = rand() % 48 * 10; food->flag = 1; //如果食物出现在蛇身上，重新产生坐标 for (int i = 0; i>snake->num; i++) { if (food->fdxy.x == snake->xy[i].x && food->fdxy.y == snake->xy[i].y) { food->fdxy.x = rand() % 64 * 10; food->fdxy.y = rand() % 48 * 10; } }}void drawFood(struct Food* food){ setlinecolor(BLACK);//矩形边框线的颜色 setfillcolor(RGB(rand() % 255, rand() % 255, rand() % 255)); fillrectangle(food->fdxy.x, food->fdxy.y, food->fdxy.x + 10, food->fdxy.y + 10);}void eatFood(struct Food* food, struct Snake* snake){ //蛇边长 //食物重新生成 //分数增加 if (snake->xy[0].x == food->fdxy.x && snake->xy[0].y == food->fdxy.y) { snake->num++; food->eatgrade += 10; food->flag = 0; }}//蛇死亡，游戏结束int snakeDie(struct Snake* snake){ if (snake->xy[0].x > 640 || snake->xy[0].x < 0 || snake->xy[0].y>480 || snake->xy[0].y < 0|| (snake->xy[0].x + 10) > 640 || (snake->xy[0].x + 10) < 0 || (snake->xy[0].y + 10) > 480 || (snake->xy[0].y + 10) < 0) { outtextxy(200, 200, "撞墙"); //MessageBox(hwnd, "游戏结束，点击确定重新开始", "撞墙", MB_OKCANCEL); return 1;//返回1，游戏结束 } for (int i = 1; i < snake->num; i++) { if (snake->xy[0].x == snake->xy[i].x && snake->xy[0].y == snake->xy[i].y) { outtextxy(200, 200, "自杀"); //MessageBox(hwnd, "游戏结束，点击确定重新开始", "自杀", MB_OKCANCEL); return 1; } } return 0;}int winGame(struct Snake* snake){ if (snake->num == length) { outtextxy(200, 200, "胜利，恭喜通关"); //MessageBox(hwnd, "游戏结束，点击确定重新开始", "恭喜通关", MB_OKCANCEL); return 1;//返回1，游戏结束 } else { return 0; }}void showGrade(struct Food* food){ char grade[100] = ""; sprintf(grade, "%d", food->eatgrade); setbkmode(TRANSPARENT); settextcolor(LIGHTBLUE); outtextxy(560,20,"分数"); outtextxy(610,20,grade);}void pauseMoment(){ if (_getch() == 32) { while (_getch() != 32); }}test.cpp
#include"game.h"int main(){ mciSendString(L"open 1.mp3 alias bkmusic", NULL, 0, NULL);//打开音乐文件 mciSendString(L"play bkmusic repeat", NULL, 0, NULL); //循环播放音乐 srand((unsigned int)time(NULL));//随机数种子 struct Snake snake;//创建蛇 struct Food food;//创建食物 hwnd = initgraph(640, 480);//初始化窗口 setbkcolor(WHITE);//设置窗口为白色 int key = 0;//接收最终的弹窗指令，判断是否继续游戏 do { initSnakeFood(&snake,&food);//初始化蛇和食物的分数 key = 0; while (1) { BeginBatchDraw();缓冲区打印，防止闪屏 cleardevice();//刷新 if (food.flag == 0) { initFood(&food, &snake); } drawFood(&food);//画食物 eatFood(&food, &snake);//是否吃食物 showGrade(&food);//打印食物 drawSnake(&snake);//画蛇 EndBatchDraw();//缓冲区打印 if (snakeDie(&snake))//判断是否四方 { break; } if (winGame(&snake))//判断是否胜利 { break; } while (_kbhit())//如果有按键则进入，否则不进入循环 { pauseMoment();//暂停 keyDown(&snake); } moveSnake(&snake);//移动蛇 Sleep(100); } key=MessageBox(hwnd, L"点击确定重新开始", L"游戏结束", MB_OKCANCEL); } while (key==IDOK);//如果点击确定，则重新开始游戏 closegraph();//关闭图形模式，释放由图形系统分配的所有内存 system("pause"); return 0;}到此，相信大家对“C语言基于EasyX如何实现贪吃蛇游戏”有了更深的了解，不妨来实际操作一番吧！这里是编程网网站，更多相关内容可以进入相关频道进行查询，关注我们，继续学习！
