banker.c

银行家算法可以解决计算机操作系统里的互斥。还有peterson等其他算法

#include<stdio.h>
#include<stdlib.h>
#define n     3
#define m     5
   int compare1(int work[],int array[])
	  {  int l,a=0;
	      for(l=0;l<n;l++)
		  {
			  if(work[l]==array[l])
			  	  a++;
			  else 
				  break;
		  }
           if(a==3)
			   return 1;
		   else return 0;
	  }	
	  int compare(int work[],int array[][n],int g)
	  {
		  int l,a=0;
	      for(l=0;l<n;l++)
		  {
			  if(work[l]>=array[g][l])
			  	  a++;
			  else 
				  break;
		  }
           if(a==3)
			   return 1;
		   else 
			   return 0;
	  }	
   int security(int available[],int need[][n],int allocation[][n],int all[])
   {
	   int work[n],i,j,w=0,r=0,l=0,finish[m]={0,0,0,0,0};
	     for(i=0;i<n;i++)
				 work[i]=available[i];
	   while(!w)
			 {   
				 for(i=0;i<m;)
				 {
					 if(!finish[i]&&compare(work,need,i))
					 {
							 for(j=0;j<n;j++)
							 work[j]+=allocation[i][j];
							  finish[i]=1;
							  r++;
							  i++;
					 }
					 else
					 {
						 i++;
						 while(finish[i])
						 i++;
					 }
				 }
				 if(r==l)
				   break;
				 else
					 l=r;
			}
			 
			 if(compare1(work,all))
					 return 1;
			 else 
					 return 0;
   }
     

  void main()
	  { 
	  int  max[m][n]={7,5,3,3,2,2,9,0,2,2,2,2,4,3,3},allocation[m][n],request[n],need[m][n],available[n],all[n]={10,5,7},i,d,j;
		   char quit;
		   int q=0;
		   for(i=0;i<m;i++)
			   for(j=0;j<n;j++)
			   { allocation[i][j]=0;
                 need[i][j]=max[i][j]-allocation[i][j];
			   } 
			   for(i=0;i<n;i++)
				   available[i]=all[i];
			  printf("---------------------struction-------------------------------\n");
			  printf("    the alloction of Every process is 0 . \n now,you may input the source of some process needed source\n if you want to stop input c\n");
           do
			  {
				scanf("%d",&d);
				for(j=0;j<n;j++)
				{
					scanf("%d",&request[j]);
					if(request[j]>need[d][j]||request[j]>available[j])
					{
						printf("not enough source\n");
						exit(0);
					}
				}
				
				
			    for(j=0;j<n;j++)
				{
				  need[d][j]-=request[j];
                  allocation[d][j]+=request[j];
				  available[j]-=request[j];
				}
                 
	            if(security(available,need,allocation,all))
                {  
					printf(" request seccussfully  the source of neccesary,\n if you want to go on request sorce ,please input process number ,then the quantity of source \n"); 
					for(i=0;i<m;i++)
			        for(j=0;j<n;j++)
				  printf("%d  ",need[i][j]);	    
				i=0;
					while(i<n)
					{  if(need[d][i]==0)
					{	q++; i++;}
					   else  break;	 
					}
					if(q==n)
					{	for(i=0;i<n;i++)
						available[i]+=allocation[d][i];
					    q=0;
					}
				}

				else
				{
					printf("this need is not securious , should abandom\n if you want to go on request sorce ,please input process number ,then the quantity of source \n");
					for(j=0;j<n;j++)
					{
				      need[d][j]+=request[j];
                      allocation[d][j]-=request[j];
				      available[j]+=request[j];
					}
				}
				scanf("%c",&quit);
				}
			  while(quit!='c');
	}