C#的银行家算法的实现


using System;

namespace 银行家算法
{
	class MainClass
	{
		public class Banker {  
			static int[] available = new int[4];       //资源数  
			static int[,] max = new int[5,4];          //最大资源需求
			static int[,] allocation = new int[5,4];   //已经分配的资源
			static int[,] need = new int[5,4];         //还需要的资源
			static int[] request = new int[4];         //存放请求
			  
			int thread;  //线程号  

			//初始化各类的值  
			public void getData(){  
				Console.Write("请输入A,B,C,D四类资源的数目:");  
				//输入A,B,C三类资源数量  
				for(int i = 0; i < 4; i++){  
					available [i] = int.Parse (Console.ReadLine ());  
				}  
				//输入进程对三类资源的最大需求  
				for(int i = 0; i < 5; i++){  
					Console.WriteLine("请输入进程" + i + "对A,B,C,D三类资源的最大需求");  
					for(int j = 0; j < 4; j++){  
						max [i,j] = int.Parse (Console.ReadLine ()); 
					}  
				}  
				//输入进程分配的三类资源数  
				for(int i = 0; i < 5; i++){  
					Console.WriteLine("请输入进程" + i + "已分配的A,B,C,D三类资源数");  
					for(int j = 0; j < 4; j++){  
						allocation[i,j] = int.Parse (Console.ReadLine ());  
					}  
				}  
				//计算进程还需要的三类资源数  
				for(int i = 0; i < 5; i++){  
					for(int j = 0; j < 4; j++){
						need[i,j] = max[i,j] - allocation[i,j];  
					}  
				}  
				//重新计算available  
				for(int i = 0; i < 4; i++){  
					for(int j = 0; j < 5; j++){  
						available[i] -= allocation[j,i];  
					}  
				}  
			}

			//用户输入要申请资源的线程和申请的资源,并进行判断  
			public void getThread(){  
				Console.WriteLine("请输入申请资源的线程");  
				int thread = int.Parse(Console.ReadLine());     //线程  
				if(thread < 0 || thread > 4){  
					Console.WriteLine("该线程不存在,请重新输入");  
					getThread();
				}else{  
					this.thread = thread;  
					Console.WriteLine("请输入申请的资源(0-4)");  
					for(int i = 0; i < 4; i++){  
						request[i] = int.Parse(Console.ReadLine());  
					}  
					if(request[0] > need[thread,0] || request[1] > need[thread,1] || request[2] > need[thread,2] || request[3] > need[thread,3]){  
						Console.WriteLine(thread+"线程申请的资源超出其需要的资源,请重新输入");  
						getThread();  
					}else{  
						if(request[0] > available[0] || request[1] > available[1] || request[2] > available[2] || request[3] > available[3]){
							Console.WriteLine(thread + "线程申请的资源大于系统资源,请重新输入");  
							getThread();  
						}  
					}  
					changeData(thread);  
					if(check(thread)){  
						getThread();  
					}else{  
						recoverData(thread);  
						getThread();  
					}  

				}  
			}  

			//thread线程请求响应后,试探性分配资源  
			public void changeData(int thread){  
				for(int i = 0; i < 4; i++){  
					//重新调整系统资源数  
					available[i] -= request[i];  
					//计算各个线程拥有资源  
					allocation[thread,i] += request[i];  
					//重新计算需求  
					need[thread,i] -= request[i];  
				}         
			}

			//安全性检查为通过,分配失败时调用,恢复系统原状           
			public void recoverData(int thread){  
				for(int i = 0; i < 4; i++){  
					//重新调整系统资源数  
					available[i] += request[i];  
					//计算各个线程拥有资源  
					allocation[thread,i] -= request[i];  
					//重新计算需求  
					need[thread,i] += request[i];  
				}   
			}  

			//对线程thread安全性检查  
			public bool check(int thread){  
				bool[] finish = new bool[5];  
				int[] work = new int[4];  
				int[] queue = new int[5];   //由于存放安全队列  
				int k = 0;//安全队列下标  
				int j;  //要判断的线程  
				int i;  
				//是否分配的标志  
				for( i = 0; i < 5; i++)  
					finish[i] = false;  
				j = thread;  
				for(i = 0; i < 4; i++){  
					work[i] = available[i];  
				}  
				while(j < 5){  
					for( i = 0; i < 4; i++){    
						if(finish[j]){  
							j++;  
							break;  
						}else if(need[j,i] > work[i]){  
							//(need[j,i]+"*"+i+work[i]);  
							j++;  
							break;  
						}else if(i == 2){  
							for(int m = 0; m < 4; m++){  
								work[m] += allocation[j,m];  
							}  
							finish[j] = true;  
							queue[k] = j;  
							k++;  
							j = 0;   //从最小线程再开始判断  
						}  
					}  
				}  
				//判断是否都属于安全状态  
				for(int p = 0; p < 5; p++){  
					if(finish[p] == false){  
						Console.WriteLine("系统不安全,资源申请失败");  
						return false;  
					}  
				}  
				Console.WriteLine("资源申请成功,安全队列为:");  
				for(int q = 0; q < 5; q++){  
					Console.WriteLine (queue [q]); 
				}   
				return true;  
			}

			//输出need和available
			public void showData(){
				Console.WriteLine("need");
				for(int i = 0; i < 5; i++){  
					for(int j = 0; j < 4; j++){  
						Console.WriteLine (need [i, j] + "     ");
					}  
				}  
				Console.WriteLine("available");  
				for(int j = 0; j < 4; j++){  
					Console.WriteLine(available[j] + "     ");  
				}  
			}

			static void Main(string[] args) {  
				Banker bk = new Banker();  
				bk.getData();  
				bk.getThread();
			}

		}

	}
}

    原文作者:银行家问题
    原文地址: https://blog.csdn.net/Alina_catty/article/details/78792085
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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