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나는 교착 상태 회피를 위해 뱅커 알고리즘을 구현했지만 ..... 안전한 순서를 얻지 못하고있다 ... 누군가 내 코드에 무엇이 잘못되었는지를 말해 줄 수있다 ... .. ??? ???? ???? 다음과 같이 친절하게 안내 ..... 프로그램 코드는 다음과 같이 프로그램의교착 상태 회피를위한 뱅커 알고리즘
#include <stdio.h>
#include <stdlib.h>
int main()
{
int Max[10][10], need[10][10], alloc[10][10], avail[10], completed[10], safeSequence[10];
int p, r, i, j, process, count;
count = 0;
printf("Enter the no of processes : ");
scanf("%d", &p);
for(i = 0; i< p; i++)
completed[i] = 0;
printf("\n\nEnter the no of resources : ");
scanf("%d", &r);
printf("\n\nEnter the Max Matrix for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ", i + 1);
for(j = 0; j < r; j++)
scanf("%d", &Max[i][j]);
}
printf("\n\nEnter the allocation for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ",i + 1);
for(j = 0; j < r; j++)
scanf("%d", &alloc[i][j]);
}
printf("\n\nEnter the Available Resources : ");
for(i = 0; i < r; i++)
scanf("%d", &avail[i]);
for(i = 0; i < p; i++)
for(j = 0; j < r; j++)
need[i][j] = Max[i][j] - alloc[i][j];
do
{
printf("\n Max matrix:\tAllocation matrix:\n");
for(i = 0; i < p; i++)
{
for(j = 0; j < r; j++)
printf("%d ", Max[i][j]);
printf("\t\t");
for(j = 0; j < r; j++)
printf("%d ", alloc[i][j]);
printf("\n");
}
process = -1;
for(i = 0; i < p; i++)
{
if(completed[i] == 0)//if not completed
{
process = i ;
for(j = 0; j < r; j++)
{
if(avail[j] < need[i][j])
{
process = -1;
break;
}
}
}
if(process != -1)
break;
}
if(process != -1)
{
printf("\nProcess %d runs to completion!", process + 1);
safeSequence[count] = process + 1;
count++;
for(j = 0; j < r; j++)
{
avail[j] += alloc[process][j];
alloc[process][j] = 0;
Max[process][j] = 0;
completed[process] = 1;
}
}
}
while(count != p && process != -1);
if(count == p)
{
printf("\nThe system is in a safe state!!\n");
printf("Safe Sequence : < ");
for(i = 0; i < p; i++)
printf("%d ", safeSequence[i]);
printf(">\n");
}
else
printf("\nThe system is in an unsafe state!!");
}
출력은 다음과 같습니다
[[email protected] 111a1059]# gcc bankerssafesequence.c
[[email protected] 111a1059]# ./a.out
Enter the no of processes : 5
Enter the no of resources : 3
Enter the Max Matrix for each process :
For process 1 : 7
5
3
For process 2 : 3
2
2
For process 3 : 7
0
2
For process 4 : 2
2
2
For process 5 : 4
3
3
Enter the allocation for each process :
For process 1 : 0
1
0
For process 2 : 2
0
0
For process 3 : 3
0
2
For process 4 : 2
1
1
For process 5 : 0
0
2
Enter the Available Resources : 3
3
2
Max matrix: Allocation matrix:
7 5 3 0 1 0
3 2 2 2 0 0
7 0 2 3 0 2
2 2 2 2 1 1
4 3 3 0 0 2
Process 2 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
7 0 2 3 0 2
2 2 2 2 1 1
4 3 3 0 0 2
Process 3 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
2 2 2 2 1 1
4 3 3 0 0 2
Process 4 runs to completion!
Max matrix: Allocation matrix:
7 5 3 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
4 3 3 0 0 2
Process 1 runs to completion!
Max matrix: Allocation matrix:
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
4 3 3 0 0 2
Process 5 runs to completion!
The system is in a safe state!!
Safe Sequence : < 2 3 4 1 5 >
버그가있는 곳을 찾기 위해 더 많은 메시지가 실행될 때이를 인쇄하도록 설정해야합니다. 숙제에 행운을 빈다. 괜찮은 표를 얻으려고하는 것처럼 보입니다. :) – usumoio
@IamJohnGalt 나는 많이 시도했지만 ......... 나는 실수를 할 수 없었다. ... 도움을 요청한 ....... – chinu