實(shí)驗(yàn)四、 死鎖檢測(cè)算法的實(shí)現(xiàn) 

實(shí)驗(yàn)?zāi)康模赫莆浙y行家算法 
實(shí)驗(yàn)?zāi)夸? /home/b32/b32019/huangdongkai/experiment4 
實(shí)驗(yàn)內(nèi)容：假定某系統(tǒng)有10類資源，編程實(shí)現(xiàn)死鎖的檢測(cè)算法。 
1） 分別讀入5個(gè)進(jìn)程的資源總需求，以及各個(gè)進(jìn)程現(xiàn)占有的資源情況； 
2） 生成系統(tǒng)可用資源向量； 
3） 判斷此狀況下系統(tǒng)是否安全，如安全打印出安全序列。 

為了方便，我的程序使用了A,B,C三個(gè)資源來模擬課本P97例子： 
源程序如下: (Banker.c ) 
＃i nclude 

#define PROCESS 5 
#define COURCE 3 

int findp = 0; 
int Done[5]; 
int Available[COURCE]; 
int Allocation[PROCESS][COURCE]; 
int Work[COURCE]; 
int Need[PROCESS][COURCE]; 
bool Finish[PROCESS]; 
bool done(); 

void readInit(void)//初始化 
{ 
printf("input the Allocation of each Process:\n"); 
for(int i = 0;i { 
printf("Process %d : ",i+1); 
for(int j = 0;j { 
scanf("%d",&Allocation[i][j]); 
} 
} 
printf("input the Need of each Process:\n"); 
for(int i = 0;i { 
printf("Process %d : ",i+1); 
for(int j = 0;j { 
scanf("%d",&Need[i][j]); 
} 
Finish[i] = false; 
} 
printf("input the Available array :\n"); 
for(int j = 0;j { 
scanf("%d",&Available[j]); 
} 
} 

void findProcess(int n)//查找安全序列 
{ 
int i = n; 
for(int j = 0;j<3;j++) 
{ 
Work[j] = Available[j]; 
} 
while(!done()||(n/5) <=5) 
{ 
if(Finish[i] == false&& 
Need[i][0]<=Work[0]&& 
Need[i][1]<=Work[1]&& 
Need[i][2]<=Work[2]) 
{ 
Work[0] += Need[i][0]; 
Work[1] += Need[i][1]; 
Work[2] += Need[i][2]; 
Finish[i] = true; 
Done[findp] = i; 
findp ++; 
} 
i = (i+1)%5; 
n++; 
} 
} 
bool done() 
{ 
if(Finish[0] == true 
&& Finish[1] == true 
&& Finish[2] == true 
&& Finish[3] == true 
&& Finish[4] == true 
) 
return true; 
else return false; 
} 

int * list()//返回安全序列 
{ 
int i = 0; 
while(!done()&&i<5) 
{ 
findProcess(i); 
i++; 
} 
return Done; 
} 
int main() 
{ 
readInit(); 
int *getlist; 
getlist = list(); 
printf("get the safe process list:\n");//打印安全序列 
for(int i = 0;i<5;i++) 
{ 
printf("P%d ",getlist[i]); 
} 
printf("\n"); 
return 0; 
} 


運(yùn)行結(jié)果為如下： 

input the Allocation of each Process: 
Process 1 : 0 1 0 
Process 2 : 2 0 0 
Process 3 : 3 0 2 
Process 4 : 2 1 1 
Process 5 : 0 0 2 
input the Need of each Process: 
Process 1 : 7 4 3 
Process 2 : 1 2 2 
Process 3 : 6 0 0 
Process 4 : 0 1 1 
Process 5 : 4 3 1 
input the Available array : 
3 3 2 
get the safe process list: 
P1 P3 P4 P0 P2