C语言实现俄罗斯方块

本文实例为大家分享了C语言俄罗斯方块的具体代码,供大家参考,具体内容如下

创新互联公司服务项目包括莲池网站建设、莲池网站制作、莲池网页制作以及莲池网络营销策划等。多年来,我们专注于互联网行业,利用自身积累的技术优势、行业经验、深度合作伙伴关系等,向广大中小型企业、政府机构等提供互联网行业的解决方案,莲池网站推广取得了明显的社会效益与经济效益。目前,我们服务的客户以成都为中心已经辐射到莲池省份的部分城市,未来相信会继续扩大服务区域并继续获得客户的支持与信任!

本代码运行环境是Windows下的VS2013
首先创建tetris.cpp
然后依次创建view.h以及view.cpp、model.h以及model.cpp。

代码如下:

view.h

#pragma once


#include 
void ShowBackground();
void ShowBrick();
void ShowGame();
void OnLeft();
void OnRight();
void OnUp();
void OnDown();

view.cpp

#include 
#include "view.h"
#include "model.h"
void OnLeft()
{//如果能够左移,则左移
 if (IsCanMove(g_nRow, g_nCol - 1))
 {
 g_nCol--;
 ShowGame();
 }
}

void OnRight()
{
 if (IsCanMove(g_nRow, g_nCol + 1))
 {
 g_nCol++;
 ShowGame();
 }
}

void OnUp()
{
 if (IsCanRotate())
 {
 Rotate();
 ShowGame();
 }
}

void OnDown()
{
 if (IsCanMove(g_nRow+1, g_nCol))
 {
 g_nRow++;
 ShowGame();
 }
 else
 {
 //固定方块至背景,并且产生新方块
 CombineBgBrick();
 GetNewBrick();
 //判断游戏是否结束,并给出对应提示
 }
}

void ShowGame()
{
 system("cls");
 CombineBgBrick();
 ShowBackground();
 DetachBgBrick();
}
void ShowBrick()
{
 for (size_t i = 0; i < 4; i++)
 {
 for (size_t j = 0; j < 4; j++)
 {
 if (g_chBrick[i][j] == 1)
 {
 printf("■");
 }
 }
 printf("\r\n");
 }
}

void ShowBackground()
{
 for (size_t nRow = 0; nRow < GAME_ROWS; nRow++)
 {
 for (size_t nCol = 0; nCol < GAME_COLS; nCol++)
 {
 if (g_chBackground[nRow][nCol] == 1)
 {
 printf("■");
 }
 else
 {
 printf("□");
 }
 }
 printf("\r\n");
 }
}

model.cpp

#include 
#include 
#include 
#include "model.h"


char g_chBackground[GAME_ROWS][GAME_COLS];
char g_chBrick[4][4];
int g_nShape = 0; //是长条还是方块,系数为16
int g_nRotate = 0; //朝向,系数为4
int g_nRow = 0;
int g_nCol = 0;
char g_chBrickPool[][4] = {
// 长条
1, 1, 1, 1,
0, 0, 0, 0,
0, 0, 0, 0,
0, 0, 0, 0,

1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0,

1, 1, 1, 1,
0, 0, 0, 0,
0, 0, 0, 0,
0, 0, 0, 0,

1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0,

// T形
1, 1, 1, 0,
0, 1, 0, 0,
0, 0, 0, 0,
0, 0, 0, 0,

0, 1, 0, 0,
1, 1, 0, 0,
0, 1, 0, 0,
0, 0, 0, 0,

0, 1, 0, 0,
1, 1, 1, 0,
0, 0, 0, 0,
0, 0, 0, 0,

1, 0, 0, 0,
1, 1, 0, 0,
1, 0, 0, 0,
0, 0, 0, 0,

//L形状
1, 0, 0, 0,
1, 0, 0, 0,
1, 1, 0, 0,
0, 0, 0, 0,

1, 1, 1, 0,
1, 0, 0, 0,
0, 0, 0, 0,
0, 0, 0, 0,

1, 1, 0, 0,
0, 1, 0, 0,
0, 1, 0, 0,
0, 0, 0, 0,

0, 0, 1, 0,
1, 1, 1, 0,
0, 0, 0, 0,
0, 0, 0, 0,
};

int IsCanRotate()
{
 char chNextShape[4][4] = { 0 };
 int nNextRotate = (g_nRotate + 1) % 4;
 int nPoolRows = g_nShape * 16 + nNextRotate * 4;
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 chNextShape[nRow][nCol] = g_chBrickPool[nRow + nPoolRows][nCol];
 }
 }
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 if (chNextShape[nRow][nCol] == 1)
 {
 if (g_chBackground[nRow + g_nRow][nCol + g_nCol] == 1)
 {
  return 0; //不能移动
 }
 }
 }
 }
 return 1;
}

void Rotate()
{
 g_nRotate = (g_nRotate + 1) % 4;
 int nPoolRows = g_nShape * 16 + g_nRotate*4;
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 g_chBrick[nRow][nCol] = g_chBrickPool[nRow + nPoolRows][nCol];
 }
 }
}

int IsCanMove(int nToRow, int nToCol)
{
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 if (g_chBrick[nRow][nCol] == 1)
 {
 if (g_chBackground[nRow + nToRow][nCol + nToCol] == 1)
 {
  return 0; //不能移动
 }
 }
 }
 }
 return 1;
}

void GetNewBrick()
{
 srand((unsigned)time(NULL));
 g_nRow = 0;
 g_nCol = GAME_COLS / 2 - 1;
 int nShapeCount = sizeof(g_chBrickPool) / sizeof(g_chBrickPool[0]) /16;
 g_nShape = rand() % nShapeCount;
 g_nRotate = rand() % 4;
 int nPoolRows = g_nShape * 16 + g_nRotate * 4;
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 g_chBrick[nRow][nCol] = g_chBrickPool[nRow+nPoolRows][nCol];
 }
 }
}

void DetachBgBrick()
{
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 if (g_chBrick[nRow][nCol] == 1)
 {
 g_chBackground[nRow + g_nRow][nCol + g_nCol] = 0;
 }
 }
 }
}

void CombineBgBrick()
{//组合块
 for (size_t nRow = 0; nRow < 4; nRow++)
 {
 for (size_t nCol = 0; nCol < 4; nCol++)
 {
 if (g_chBrick[nRow][nCol] == 1)
 {
 g_chBackground[nRow+g_nRow][nCol+g_nCol] = 1;
 }
 }
 }
}

void InitBackground()
{//初始化背景
 for (size_t nRow = 0; nRow < GAME_ROWS; nRow++)
 {
 for (size_t nCol = 0; nCol < GAME_COLS; nCol++)
 {
 if (nRow == GAME_ROWS - 1
 || nCol == 0
 || nCol == GAME_COLS - 1)
 {
 g_chBackground[nRow][nCol] = 1;
 }
 else
 {
 g_chBackground[nRow][nCol] = 0;
 }
 }
 }
}

model.h

#pragma once

#define GAME_ROWS 20
#define GAME_COLS 12

extern char g_chBackground[GAME_ROWS][GAME_COLS];
extern char g_chBrick[4][4];
extern int g_nRow;
extern int g_nCol;

void InitBackground();
void GetNewBrick();
void CombineBgBrick();
void DetachBgBrick();
int IsCanMove(int nToRow, int nToCol);
void Rotate();
int IsCanRotate();

tetris.cpp

#include "stdafx.h"
#include 
#include 
#include 
#include "model.h"
#include "view.h"


int main(int argc, char* argv[])
{
 InitBackground();
 GetNewBrick();
 CombineBgBrick();
 ShowBackground();
 DetachBgBrick();
 char chInput = 0;
 clock_t clkStart = clock();
 clock_t clkEnd = clock();
 while (1)
 {
 clkEnd = clock();
 if (clkEnd - clkStart > 1000)
 {
 clkStart = clkEnd;
 OnDown();
 }
 if (_kbhit() != 0)
 {
 chInput = _getch();
 }
 switch (chInput)
 {
 case 'a':
 OnLeft();
 break;
 case 'w':
 OnUp();
 break;
 case 's':
 OnDown();
 break;
 case 'd':
 OnRight();
 break;
 default:
 break;
 }
 chInput = 0;
 }
 return 0;
}

更多关于俄罗斯方块的文章,请点击查看专题:《俄罗斯方块》

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持创新互联。


本文名称:C语言实现俄罗斯方块
转载来于:http://hbruida.cn/article/jgscid.html