可可.txt

用位示图进行文件空间管理,操作系统实验.

// sdf.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream.h>

//#define null 0

struct File
{
	char name;
	int len;
	int blockNum;
	File * next;
};

int M,N,K;
int * chart;
File * head;//,* tail;
void init();
void create();
void del();
void display();

int main(int argc, char* argv[])
{
	cout << "输入柱面数，磁道数，扇区数：";
	cin >> M >> N >> K;
	chart = new int[M*N*K];
	for(int i = 0; i < M*N*K; i++)
		chart[i] = 0;
	init();
	int cc;
	cout << "请选择：\n1.分配2.删除3.显示0.退出\n";
	cin >> cc;
	while(cc)
	{
		switch(cc)
		{
			case 1:create();display();break;
			case 2:del();display();break;
			case 3:display();break;
			case 0:break;
		}
		cout << "请选择：\n1.分配2.删除3.显示0.退出\n";
		cin >> cc;
	}
	return 0;
}

void init()
{
	head = new File;//tail = new File;
	(*head).next = NULL;//(*tail).next = null;
}

void create()
{
	File * temp = new File;
	cout << "输入新建文件名及长度：";
	cin >> (*temp).name >> (*temp).len ;
	(*temp).next = NULL;
	int num = 0;
	for(int i = 0; i < M*N*K;)
	{
		if(chart[i++] == 0)
		{
			num ++;
			if(num == (*temp).len)
			{
				(*temp).blockNum = i - num;
				//File * p = &temp;
				for(int j = 0; j < (*temp).len; j++)
				{
					chart[(*temp).blockNum + j] = 1;
				}
				if((*head).next != NULL)
				{
					(*temp).next = (*head).next;
					(*head).next = temp;
					//(*tail).next = p;
					//tail = p;
					return;
				}
				else
				{
					(*head).next = temp;
					return;
				}
			}
		}
		else
		{
			num = 0;
		}
	}
	cout << "分配失败...........\n";
}

void del()
{
	char ch;
	cout << "输入要删除的文件名：";
	cin >> ch;
	File * temp = (*head).next;
	File * p = head;
	cout << "输入要删除的文件名：";
	while(temp != NULL)
	{
		if((*temp).name == ch)
		{
			for(int i = 0; i < (*temp).len;i++)
			{
				chart[(*temp).blockNum + i] = 0;
			}
			(*p).next = (*temp).next;
			//delete temp;
			return;
		}
		else 
		{
			p = temp;
			temp = (*temp).next;
		}
	}
}

void display()
{
	cout << "位示图:\n";
	for(int i = 0; i < M; i++)
	{
		for(int j = 0; j < M*K; j++)
		{
			cout << chart[i*M*K + j];
		}
		cout << endl;
	}
}