銀行家演算法c語言代碼

⑴ 求n個數的全排列，n不定。用c語言。用於銀行家演算法中求安全序列

好久沒用c了，所以代碼可能要用到偽代碼
先定義a[maxn]
用子函數遞歸
void p(int x)
{
if (n == x+1)
{
//foreach a print
//輸出數組a
}
for (int i=1 to n)
{
a[x] = i;
p(x+1);
a[x] = 0;
}
}
主函數main調用p(n)

⑵ 怎樣用C語言實現銀行家演算法

#include<stdio.h>
struct claim
{
int user;
int num[3];
}claims;
int input()
{
printf("please input your request:user(0~4):\n");
scanf("%d",&claims.user);
printf("input the number of resource a:\n");
scanf("%d",&claims.num[0]);
printf("input the number of resource b:\n");
scanf("%d",&claims.num[1]);
printf("input the number of resource c:\n");
scanf("%d",&claims.num[2]);
return 1;
}
int safety_chk(int alloc[][3],int need[][3],int avail[3])
{
int work[3],finish[5];
for(int p=0;p<5;p++)//i大於2後對WORK是無意義的
{
work[p]=avail[p];
finish[p]=0;
}
for(int i=0;i<5;i++)
{
if(finish[i]==0&&
need[i][0]<=work[0]&&
need[i][1]<=work[1]&&
need[i][2]<=work[2] )
{
for(int j=0;j<3;j++)
work[j]=alloc[i][j]+work[j];
finish[i]=1;
i=-1;//重頭再來
}
}
for(i=0;i<5;i++)
{
if(finish[i]==0)
return 0;
}
return 1;
}

int process(int alloc[][3],int need[][3],int avail[3])
{
int ret;
input();
for(int i=0;i<3;i++) //out of resource number
{
if(claims.num[i]>need[claims.user][i]||claims.num[i]>avail[i])
return 0;
}
for(i=0;i<3;i++)//trying
{
avail[i]=avail[i]-claims.num[i];
alloc[claims.user][i]=alloc[claims.user][i]+claims.num[i];
need[claims.user][i]=need[claims.user][i]-claims.num[i];
}
if((ret=safety_chk(alloc,need,avail)==0))
{

printf("safety_chk's result %d \n",0);
for(i=0;i<3;i++)
{
avail[i]=avail[i]+claims.num[i];
alloc[claims.user][i]=alloc[claims.user][i]-claims.num[i];
need[claims.user][i]=need[claims.user][i]+claims.num[i];
}
return 0;
}
else
{

printf("safety_chk's result %d \n",1);
}
return 1;
}
void main()
{
int alloc[5][3]={{0,1,0},{2,0,0},{3,0,2},{2,1,1},{0,0,2}};
int need[5][3]={{7,4,3},{1,2,2},{6,0,0},{0,1,1},{4,3,1}};
int avail[3]={3,3,2};
if(process(alloc,need,avail)==0)
printf("sorry,we cannot help you!\n");
else printf("operation complete!\n");
return;
}

⑶ 銀行家演算法的C語言程序

1.根據下面給出的系統中資源分配情況，以及各個進程的資源申請情況，通過銀行家演算法來判斷各進程的資源請求能否滿足（要求記錄程序的運行過程）。 已分配的

⑷ 演算法上機實驗如圖所示，用c語言實現

#include<stdio.h>

#include<stdlib.h>

struct node {

int data;//數據域

struct node *R;//左孩子

struct node *L;//右孩子

};

int a[] = {1, 2, 3, -1, -1, -1, 4, 5,7,-1,-1,-1,6,-1,-1};//二叉樹序列

int k = 0;

node* buildTree(node *tree) {//創建二叉樹

int data=a[k++];//

if (data== - 1) {//-1代表空結點

tree = NULL;

}

else {//非空結點

tree = (node*)malloc(sizeof(node));//分配內存

tree->data = data;//數據域賦值

tree->L = tree->R = NULL;//左右孩子賦空

tree->L=buildTree(tree->L);//前往左孩子

tree->R=buildTree(tree->R);//前往右孩子

}

return tree;//返回根結點地址

}

void dfs1(node *tree) {//前序遍歷

if (tree) {

printf("%d ", tree->data);

dfs1(tree->L);

dfs1(tree->R);

}

}

void dfs2(node *tree) {//中序

if (tree) {

dfs2(tree->L);

printf("%d ", tree->data);

dfs2(tree->R);

}

}

void dfs3(node *tree) {//後序

if (tree) {

dfs3(tree->L);

dfs3(tree->R);

printf("%d ", tree->data);

}

}

void level(node *tree){

//層次遍歷，類似與bfs(廣度優先搜索)

//需要一個隊列作為輔助數據結構

node* q[100];//隊列

int f=0,r=0;//頭，尾指針

q[r++]=tree;//根結點入隊

while(f!=r){

node *t=q[f++];//出隊

printf("%d ",t->data);//輸出

if(t->L!=NULL){//非空左孩子入隊

q[r++]=t->L;

}

if(t->R!=NULL){//非空右孩子入隊

q[r++]=t->R;

}

}

}

int count(node *tree){

if(tree==NULL){

return 0;

}

else{

int n,m;

n=count(tree->L);//左子樹結點個數

m=count(tree->R);//右子樹結點個數

return n+m+1;//返回左右子樹結點個數之和

}

}

int main() {

node *tree = NULL;

tree=buildTree(tree);

printf(" 前序遍歷： ");

dfs1(tree);

printf(" 中序遍歷： ");

dfs2(tree);

printf(" 後序遍歷： ");

dfs3(tree);

printf(" 層次遍歷： ");

level(tree);

printf(" 二叉樹結點個數：%d",count(tree));

return 0;

}

⑸ 改程序——銀行家演算法C語言版

你是學軟體的學生嗎 ?
我是學軟體的，我在操作系統里學了銀行家演算法，等我看了,在告訴你我的看法,好嗎？
願意的話希望交流一下 我的郵箱[email protected]

⑹ c語言銀行家演算法安全性判別

把1作為參數傳給yanzheng（） yanzheng（int m)

然後驗證函數里修改：

work=Avaliable;
i=m;
while(i<m)
{
if(Finish[i]==false&&Need[i]<=work)
{
work=work+Allocation[i];
Finish[i]=true;
anquan[k]=i;
k++;
i=0;
}
else
i++;
}

⑺ 關於銀行家演算法中部分代碼不理解，請高手幫忙指點下謝謝！！！

lass ThreadTest {
static int type = 4, num = 10; //定義資源數目和線程數目
static int[] resource = new int[type]; //系統資源總數
//static int[] Resource = new int[type]; //副本
static Random rand = new Random();
static Bank[] bank = new Bank[num]; //線程組
Bank temp = new Bank();

public void init() {
//初始化組中每個線程，隨機填充系統資源總數
for(int i = 0; i < type; i++)
resource[i] = rand.nextInt(10) + 80;
System.out.print("Resource:");
for(int i = 0; i < type; i++)
System.out.print(" " + resource[i]);
System.out.println("");
for(int i = 0; i < bank.length; i++)
bank[i] = new Bank("#" + i);
}
public ThreadTest4() {
init();
}

class Bank extends Thread {
//銀行家演算法避免死鎖
public int[]
max = new int[type], //總共需求量
need = new int[type], //尚需資源量
allocation = new int[type]; //已分配量
private int[]
request = new int[type], //申請資源量
Resource = new int[type]; //資源副本
private boolean isFinish = false; //線程是否完成
int[][] table = new int[bank.length][type*4]; //二維資源分配表

private void init() {
// 隨機填充總共、尚需、已分配量
synchronized(resource) {
for(int i = 0; i < type; i++) {
max[i] = rand.nextInt(5) + 10;
need[i] = rand.nextInt(10);
allocation[i] = max[i] - need[i];
resource[i] -= allocation[i]; //從系統資源中減去已分配的
}
printer();
for(int i = 0; i < type; i++) {
if(resource[i] < 0) {
//若出現已分配量超出系統資源總數的錯誤則退出
System.out.println("The summation of Threads' allocations is out of range!");
System.exit(1);
}
}
}
}

public Bank(String s) {
setName(s);
init();
start();
}
public Bank() {
//none
}

public void run() {
try {
sleep(rand.nextInt(2000));
}
catch(InterruptedException e) {
throw new RuntimeException(e);
}
while(true) {
//程序沒有完成時一直不斷申請資源
if(askFor() == false) {
try {
sleep(1000);
}
catch(InterruptedException e) {
throw new RuntimeException(e);
}
}
else
tryRequest();
if(noNeed() == true)
break;
}
//休眠一段時間模擬程序運行
try {
sleep(1000);
}
catch(InterruptedException e) {
throw new RuntimeException(e);
}
System.out.println(getName() + " finish!");
synchronized(resource) {
//運行結束釋放佔有資源
for(int i = 0; i < type; i++) {
resource[i] += allocation[i];
need[i] = allocation[i] = max[i] = 0;
}
}
}

private void printer() {
//列印當前資源信息
System.out.print(getName() + " Max:");
for(int i = 0; i < type; i++)
System.out.print(" " + max[i]);
System.out.print(" Allocation:");
for(int i = 0; i < type; i++)
System.out.print(" " + allocation[i]);
System.out.print(" Need:");
for(int i = 0; i < type; i++)
System.out.print(" " + need[i]);
System.out.print(" Available:");
for(int i = 0; i < type; i++)
System.out.print(" " + resource[i]);
System.out.println("");
}
private boolean askFor() {
//隨機產生申請資源量並檢測是否超標
boolean canAsk = false;
for(int i = 0; i < type; i++) {
request[i] = rand.nextInt(20);
//防止申請量超過所需量
if(request[i] > need[i])
request[i] = need[i];
}
for(int i = 0; i < type; i++) //防止隨機申請資源全為0
if(request[i] > 0)
canAsk = true;
synchronized(resource) {
//鎖住可供資源檢查是否超標
for(int i = 0; i < type; i++) {
if(request[i] > resource[i])
//如果申請資源超過可供資源則等待一段時間後重新申請
return false;
}
}
return canAsk;
}
private void tryRequest() {
//創建副本嘗試分配請求
synchronized(resource) {
for(int i = 0; i < type; i++)
//依然要防止請求量超出范圍
if(request[i] > resource[i])
return;
for(int i = 0; i < type; i++) {
//復制資源量並減去需求量到一個副本上
Resource[i] = resource[i];
Resource[i] -= request[i];
}
System.out.print(getName() + " ask for:");
for(int i = 0; i < type; i++)
System.out.print(" " + request[i]);
System.out.println("");
if(checkSafe() == true) {
//如果檢查安全則將副本值賦給資源量並修改佔有量和需求量
for(int i = 0; i < type; i++) {
resource[i] = Resource[i];
allocation[i] += request[i];
need[i] -= request[i];
}
System.out.println(getName() + " request succeed!");
}
else
System.out.println(getName() + " request fail!");
}
}
private boolean checkSafe() {
//銀行家演算法檢查安全性
synchronized(bank) {
//將線程資源信息放入二維資源分配表檢查安全性，0~type可用資源／type~type*2所需資源／type*2~type*3佔有資源／type*3~-1可用+佔用資源
for(int i = 0; i < bank.length; i++) {
for(int j = type; j < type*2; j++) {
table[i][j] = bank[i].need[j%type];
}
for(int j = type*2; j < type*3; j++) {
table[i][j] = bank[i].allocation[j%type];
}
}
//冒泡排序按需求資源從小到大排
for(int i = 0; i < bank.length; i++) {
for(int j = i; j < bank.length-1; j++) {
sort(j, 4);
}
}
//進行此時刻的安全性檢查
for(int i = 0; i < type; i++) {
table[0][i] = Resource[i];
table[0][i+type*3] = table[0][i] + table[0][i+type*2];
if(table[0][i+type*3] < table[1][i+type])
return false;
}
for(int j = 1; j < bank.length-1; j++) {
for(int k = 0; k < type; k++) {
table[j][k] = table[j-1][k+type*3];
table[j][k+type*3] = table[j][k] + table[j][k+type*2];
if(table[j][k+type*3] < table[j+1][k+type])
return false;
}
}
}
return true;
}
private void sort(int j, int k) {
//遞歸冒泡排序
int tempNum;
if(table[j][k] > table[j+1][k]) {
for(int i = type; i < type*2; i++) {
tempNum = table[j][i];
table[j][i] = table[j+1][i];
table[j+1][i] = tempNum;
}
/*temp = bank[j];
bank[j] = bank[j+1];
bank[j+1] = temp;*/
}
else if(table[j][k] == table[j+1][k] && k < type*2) //此資源量相同時遞歸下一個資源量排序並且防止超出范圍
sort(j, k+1);
}
private boolean noNeed() {
//是否還需要資源
boolean finish = true;
for(int i = 0; i < type; i++) {
if(need[i] != 0) {
finish = false;
break;
}
}
return finish;
}
}

public static void main(String[] args) {
ThreadTest t = new ThreadTest();
//後台線程，設定程序運行多長時間後自動結束
new Timeout(30000, "---Stop!!!---");
}
}

⑻ 高分求銀行家演算法c語言版

#include "malloc.h"
#include "stdio.h"
#include "stdlib.h"
#define alloclen sizeof(struct allocation)
#define maxlen sizeof(struct max)
#define avalen sizeof(struct available)
#define needlen sizeof(struct need)
#define finilen sizeof(struct finish)
#define pathlen sizeof(struct path)
struct allocation
{
int value;
struct allocation *next;
};
struct max
{
int value;
struct max *next;
};
struct available /*可用資源數*/
{
int value;
struct available *next;
};
struct need /*需求資源數*/
{
int value;
struct need *next;
};
struct path
{
int value;
struct path *next;
};
struct finish
{
int stat;
struct finish *next;
};
int main()
{
int row,colum,status=0,i,j,t,temp,processtest;
struct allocation *allochead,*alloc1,*alloc2,*alloctemp;
struct max *maxhead,*maxium1,*maxium2,*maxtemp;
struct available *avahead,*available1,*available2,*workhead,*work1,*work2,*worktemp,*worktemp1;
struct need *needhead,*need1,*need2,*needtemp;
struct finish *finihead,*finish1,*finish2,*finishtemp;
struct path *pathhead,*path1,*path2;
printf("\n請輸入系統資源的種類數:");
scanf("%d",&colum);
printf("請輸入現時內存中的進程數:");
scanf("%d",&row);
printf("請輸入已分配資源矩陣:\n");
for(i=0;i<row;i++)
{
for (j=0;j<colum;j++)
{
printf("請輸入已分配給進程 p%d 的 %c 種系統資源:",i,'A'+j);
if(status==0)
{
allochead=alloc1=alloc2=(struct allocation*)malloc(alloclen);
alloc1->next=alloc2->next=NULL;
scanf("%d",&allochead->value);
status++;
}
else
{
alloc2=(struct allocation *)malloc(alloclen);
scanf("%d,%d",&alloc2->value);
if(status==1)
{
allochead->next=alloc2;
status++;
}
alloc1->next=alloc2;
alloc1=alloc2;
}
}
}
alloc2->next=NULL;
status=0;
printf("請輸入最大需求矩陣:\n");
for(i=0;i<row;i++)
{
for (j=0;j<colum;j++)
{
printf("請輸入進程 p%d 種類 %c 系統資源最大需求:",i,'A'+j);
if(status==0)
{
maxhead=maxium1=maxium2=(struct max*)malloc(maxlen);
maxium1->next=maxium2->next=NULL;
scanf("%d",&maxium1->value);
status++;
}
else
{
maxium2=(struct max *)malloc(maxlen);
scanf("%d,%d",&maxium2->value);
if(status==1)
{
maxhead->next=maxium2;
status++;
}
maxium1->next=maxium2;
maxium1=maxium2;
}
}
}
maxium2->next=NULL;
status=0;
printf("請輸入現時系統剩餘的資源矩陣:\n");
for (j=0;j<colum;j++)
{
printf("種類 %c 的系統資源剩餘:",'A'+j);
if(status==0)
{
avahead=available1=available2=(struct available*)malloc(avalen);
workhead=work1=work2=(struct available*)malloc(avalen);
available1->next=available2->next=NULL;
work1->next=work2->next=NULL;
scanf("%d",&available1->value);
work1->value=available1->value;
status++;
}
else
{
available2=(struct available*)malloc(avalen);
work2=(struct available*)malloc(avalen);
scanf("%d,%d",&available2->value);
work2->value=available2->value;
if(status==1)
{
avahead->next=available2;
workhead->next=work2;
status++;
}
available1->next=available2;
available1=available2;
work1->next=work2;
work1=work2;
}
}
available2->next=NULL;
work2->next=NULL;
status=0;
alloctemp=allochead;
maxtemp=maxhead;
for(i=0;i<row;i++)
for (j=0;j<colum;j++)
{
if(status==0)
{
needhead=need1=need2=(struct need*)malloc(needlen);
need1->next=need2->next=NULL;
need1->value=maxtemp->value-alloctemp->value;
status++;
}
else
{
need2=(struct need *)malloc(needlen);
need2->value=(maxtemp->value)-(alloctemp->value);
if(status==1)
{
needhead->next=need2;
status++;
}
need1->next=need2;
need1=need2;
}
maxtemp=maxtemp->next;
alloctemp=alloctemp->next;
}
need2->next=NULL;
status=0;
for(i=0;i<row;i++)
{
if(status==0)
{
finihead=finish1=finish2=(struct finish*)malloc(finilen);
finish1->next=finish2->next=NULL;
finish1->stat=0;
status++;
}
else
{
finish2=(struct finish*)malloc(finilen);
finish2->stat=0;
if(status==1)
{
finihead->next=finish2;
status++;
}
finish1->next=finish2;
finish1=finish2;
}
}
finish2->next=NULL; /*Initialization compleated*/
status=0;
processtest=0;
for(temp=0;temp<row;temp++)
{
alloctemp=allochead;
needtemp=needhead;
finishtemp=finihead;
worktemp=workhead;
for(i=0;i<row;i++)
{
worktemp1=worktemp;
if(finishtemp->stat==0)
{
for(j=0;j<colum;j++,needtemp=needtemp->next,worktemp=worktemp->next)
if(needtemp->value<=worktemp->value)
processtest++;
if(processtest==colum)
{
for(j=0;j<colum;j++)
{
worktemp1->value+=alloctemp->value;
worktemp1=worktemp1->next;
alloctemp=alloctemp->next;
}
if(status==0)
{
pathhead=path1=path2=(struct path*)malloc(pathlen);
path1->next=path2->next=NULL;
path1->value=i;
status++;
}
else
{
path2=(struct path*)malloc(pathlen);
path2->value=i;
if(status==1)
{
pathhead->next=path2;
status++;
}
path1->next=path2;
path1=path2;
}
finishtemp->stat=1;
}
else
{
for(t=0;t<colum;t++)
alloctemp=alloctemp->next;
finishtemp->stat=0;
}
}
else
for(t=0;t<colum;t++)
{
needtemp=needtemp->next;
alloctemp=alloctemp->next;
}
processtest=0;
worktemp=workhead;
finishtemp=finishtemp->next;
}
}
path2->next=NULL;
finishtemp=finihead;
for(temp=0;temp<row;temp++)
{
if(finishtemp->stat==0)
{
printf("\n系統處於非安全狀態!\n");
exit(0);
}
finishtemp=finishtemp->next;
}
printf("\n系統處於安全狀態.\n");
printf("\n安全序列為: \n");
do
{
printf("p%d ",pathhead->value);
}
while(pathhead=pathhead->next);
printf("\n");
return 0;
}

⑼ 銀行家演算法的演算法實現

在避免死鎖的方法中，所施加的限制條件較弱，有可能獲得令人滿意的系統性能。在該方法中把系統的狀態分為安全狀態和不安全狀態，只要能使系統始終都處於安全狀態，便可以避免發生死鎖。
銀行家演算法的基本思想是分配資源之前，判斷系統是否是安全的；若是，才分配。它是最具有代表性的避免死鎖的演算法。
設進程cusneed提出請求REQUEST [i]，則銀行家演算法按如下規則進行判斷。
(1)如果REQUEST [cusneed] [i]<= NEED[cusneed][i]，則轉（2)；否則，出錯。
(2)如果REQUEST [cusneed] [i]<= AVAILABLE[i]，則轉（3)；否則，等待。
(3)系統試探分配資源，修改相關數據：
AVAILABLE[i]-=REQUEST[cusneed][i];
ALLOCATION[cusneed][i]+=REQUEST[cusneed][i];
NEED[cusneed][i]-=REQUEST[cusneed][i];
(4)系統執行安全性檢查，如安全，則分配成立；否則試探險性分配作廢，系統恢復原狀，進程等待。 （1)設置兩個工作向量Work=AVAILABLE;FINISH
(2)從進程集合中找到一個滿足下述條件的進程，
FINISH==false;
NEED<=Work;
如找到，執行（3)；否則，執行（4)
(3)設進程獲得資源，可順利執行，直至完成，從而釋放資源。
Work=Work+ALLOCATION;
Finish=true;
GOTO 2
(4)如所有的進程Finish= true，則表示安全；否則系統不安全。
銀行家演算法流程圖
演算法（C語言實現） #include<STRING.H>#include<stdio.h>#include<stdlib.h>#include<CONIO.H>/*用到了getch()*/#defineM5/*進程數*/#defineN3/*資源數*/#defineFALSE0#defineTRUE1/*M個進程對N類資源最大資源需求量*/intMAX[M][N]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}};/*系統可用資源數*/intAVAILABLE[N]={10,5,7};/*M個進程已分配到的N類數量*/intALLOCATION[M][N]={{0,0,0},{0,0,0},{0,0,0},{0,0,0},{0,0,0}};/*M個進程已經得到N類資源的資源量*/intNEED[M][N]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}};/*M個進程還需要N類資源的資源量*/intRequest[N]={0,0,0};voidmain(){inti=0,j=0;charflag;voidshowdata();voidchangdata(int);voidrstordata(int);intchkerr();showdata();enter:{printf(請輸入需申請資源的進程號（從0到);printf(%d,M-1);printf(）:);scanf(%d,&i);}if(i<0||i>=M){printf(輸入的進程號不存在，重新輸入! );gotoenter;}err:{printf(請輸入進程);printf(%d,i);printf(申請的資源數 );printf(類別:ABC );printf();for(j=0;j<N;j++){scanf(%d,&Request[j]);if(Request[j]>NEED[i][j]){printf(%d,i);printf(號進程);printf(申請的資源數>進程);printf(%d,i);printf(還需要);printf(%d,j);printf(類資源的資源量!申請不合理，出錯!請重新選擇! );gotoerr;}else{if(Request[j]>AVAILABLE[j]){printf(進程);printf(%d,i);printf(申請的資源數大於系統可用);printf(%d,j);printf(類資源的資源量!申請不合理，出錯!請重新選擇! );gotoerr;}}}}changdata(i);if(chkerr()){rstordata(i);showdata();}elseshowdata();printf( );printf(按'y'或'Y'鍵繼續,否則退出 );flag=getch();if(flag=='y'||flag=='Y'){gotoenter;}else{exit(0);}}/*顯示數組*/voidshowdata(){inti,j;printf(系統可用資源向量: );printf(***Available*** );printf(資源類別:ABC );printf(資源數目:);for(j=0;j<N;j++){printf(%d,AVAILABLE[j]);}printf( );printf( );printf(各進程還需要的資源量: );printf(******Need****** );printf(資源類別:ABC );for(i=0;i<M;i++){printf();printf(%d,i);printf(號進程:);for(j=0;j<N;j++){printf(%d,NEED[i][j]);}printf( );}printf( );printf(各進程已經得到的資源量: );printf(***Allocation*** );printf(資源類別:ABC );for(i=0;i<M;i++){printf();printf(%d,i);printf(號進程:);/*printf(: );*/for(j=0;j<N;j++){printf(%d,ALLOCATION[i][j]);}printf( );}printf( );}/*系統對進程請求響應，資源向量改變*/voidchangdata(intk){intj;for(j=0;j<N;j++){AVAILABLE[j]=AVAILABLE[j]-Request[j];ALLOCATION[k][j]=ALLOCATION[k][j]+Request[j];NEED[k][j]=NEED[k][j]-Request[j];}}/*資源向量改變*/voidrstordata(intk){intj;for(j=0;j<N;j++){AVAILABLE[j]=AVAILABLE[j]+Request[j];ALLOCATION[k][j]=ALLOCATION[k][j]-Request[j];NEED[k][j]=NEED[k][j]+Request[j];}}/*安全性檢查函數*/intchkerr()//在假定分配資源的情況下檢查系統的安全性{intWORK[N],FINISH[M],temp[M];//temp[]用來記錄進程安全執行的順序inti,j,m,k=0,count;for(i=0;i<M;i++)FINISH[i]=FALSE;for(j=0;j<N;j++)WORK[j]=AVAILABLE[j];//把可利用資源數賦給WORK[]for(i=0;i<M;i++){count=0;for(j=0;j<N;j++)if(FINISH[i]==FALSE&&NEED[i][j]<=WORK[j])count++;if(count==N)//當進程各類資源都滿足NEED<=WORK時{for(m=0;m<N;m++)WORK[m]=WORK[m]+ALLOCATION[i][m];FINISH[i]=TRUE;temp[k]=i;//記錄下滿足條件的進程k++;i=-1;}}for(i=0;i<M;i++)if(FINISH[i]==FALSE){printf(系統不安全!!!本次資源申請不成功!!! );return1;}printf( );printf(經安全性檢查，系統安全，本次分配成功。 );printf( );printf(本次安全序列：);for(i=0;i<M;i++)//列印安全系統的進程調用順序{printf(進程);printf(%d,temp[i]);if(i<M-1)printf(->);}printf( );return0;}

⑽ 急！銀行家演算法用C語言編寫．全部程序．

銀行家演算法
銀行家演算法是一種最有代表性的避免死鎖的演算法。
要解釋銀行家演算法，必須先解釋操作系統安全狀態和不安全狀態。
安全狀態：如果存在一個由系統中所有進程構成的安全序列P1，…，Pn，則系統處於安全狀態。安全狀態一定是沒有死鎖發生。
不安全狀態:不存在一個安全序列。不安全狀態不一定導致死鎖。
那麼什麼是安全序列呢？
安全序列：一個進程序列{P1，…，Pn}是安全的，如果對於每一個進程Pi(1≤i≤n），它以後尚需要的資源量不超過系統當前剩餘資源量與所有進程Pj (j < i )當前佔有資源量之和。
銀行家演算法：
我們可以把操作系統看作是銀行家，操作系統管理的資源相當於銀行家管理的資金，進程向操作系統請求分配資源相當於用戶向銀行家貸款。操作系統按照銀行家制定的規則為進程分配資源，當進程首次申請資源時，要測試該進程對資源的最大需求量，如果系統現存的資源可以滿足它的最大需求量則按當前的申請量分配資源，否則就推遲分配。當進程在執行中繼續申請資源時，先測試該進程已佔用的資源數與本次申請的資源數之和是否超過了該進程對資源的最大需求量。若超過則拒絕分配資源，若沒有超過則再測試系統現存的資源能否滿足該進程尚需的最大資源量，若能滿足則按當前的申請量分配資源，否則也要推遲分配。
演算法：
n：系統中進程的總數
m：資源類總數
Available: ARRAY[1..m] of integer;
Max: ARRAY[1..n,1..m] of integer;
Allocation: ARRAY[1..n,1..m] of integer;
Need: ARRAY[1..n,1..m] of integer;
Request: ARRAY[1..n,1..m] of integer;
符號說明：
Available 可用剩餘資源
Max 最大需求
Allocation 已分配資源
Need 需求資源
Request 請求資源
當進程pi提出資源申請時，系統執行下列
步驟：(「=」為賦值符號，「==」為等號)
step（1）若Request<=Need, goto step（2）；否則錯誤返回
step（2）若Request<=Available, goto step（3）；否則進程等待
step（3）假設系統分配了資源，則有：
Available=Available-Request;
Allocation=Allocation+Request;
Need=Need-Request
若系統新狀態是安全的，則分配完成
若系統新狀態是不安全的，則恢復原狀態，進程等待
為進行安全性檢查，定義數據結構：
Work:ARRAY[1..m] of integer;
Finish:ARRAY[1..n] of Boolean;
安全性檢查的步驟：
step (1):
Work=Available;
Finish=false;
step (2) 尋找滿足條件的i：
a.Finish==false;
b.Need<=Work;
如果不存在，goto step(4)
step(3)
Work=Work+Allocation;
Finish=true;
goto step(2)
step (4) 若對所有i,Finish=true,則系統處於安全狀態，否則處於不安全狀態
/* 銀行家演算法，操作系統概念(OS concepts Six Edition)
reedit by Johnny hagen，SCAU，run at vc6.0
*/
#include "malloc.h"
#include "stdio.h"
#include "stdlib.h"
#define alloclen sizeof(struct allocation)
#define maxlen sizeof(struct max)
#define avalen sizeof(struct available)
#define needlen sizeof(struct need)
#define finilen sizeof(struct finish)
#define pathlen sizeof(struct path)
struct allocation
{
int value;
struct allocation *next;
};
struct max
{
int value;
struct max *next;
};
struct available /*可用資源數*/
{
int value;
struct available *next;
};
struct need /*需求資源數*/
{
int value;
struct need *next;
};
struct path
{
int value;
struct path *next;
};
struct finish
{
int stat;
struct finish *next;
};
int main()
{
int row,colum,status=0,i,j,t,temp,processtest;
struct allocation *allochead,*alloc1,*alloc2,*alloctemp;
struct max *maxhead,*maxium1,*maxium2,*maxtemp;
struct available *avahead,*available1,*available2,*workhead,*work1,*work2,*worktemp,*worktemp1;
struct need *needhead,*need1,*need2,*needtemp;
struct finish *finihead,*finish1,*finish2,*finishtemp;
struct path *pathhead,*path1,*path2;
printf("\n請輸入系統資源的種類數:");
scanf("%d",&colum);
printf("請輸入現時內存中的進程數:");
scanf("%d",&row);
printf("請輸入已分配資源矩陣:\n");
for(i=0;i<row;i++)
{
for (j=0;j<colum;j++)
{
printf("請輸入已分配給進程 p%d 的 %c 種系統資源:",i,'A'+j);
if(status==0)
{
allochead=alloc1=alloc2=(struct allocation*)malloc(alloclen);
alloc1->next=alloc2->next=NULL;
scanf("%d",&allochead->value);
status++;
}
else
{
alloc2=(struct allocation *)malloc(alloclen);
scanf("%d,%d",&alloc2->value);
if(status==1)
{
allochead->next=alloc2;
status++;
}
alloc1->next=alloc2;
alloc1=alloc2;
}
}
}
alloc2->next=NULL;
status=0;
printf("請輸入最大需求矩陣:\n");
for(i=0;i<row;i++)
{
for (j=0;j<colum;j++)
{
printf("請輸入進程 p%d 種類 %c 系統資源最大需求:",i,'A'+j);
if(status==0)
{
maxhead=maxium1=maxium2=(struct max*)malloc(maxlen);
maxium1->next=maxium2->next=NULL;
scanf("%d",&maxium1->value);
status++;
}
else
{
maxium2=(struct max *)malloc(maxlen);
scanf("%d,%d",&maxium2->value);
if(status==1)
{
maxhead->next=maxium2;
status++;
}
maxium1->next=maxium2;
maxium1=maxium2;
}
}
}
maxium2->next=NULL;
status=0;
printf("請輸入現時系統剩餘的資源矩陣:\n");
for (j=0;j<colum;j++)
{
printf("種類 %c 的系統資源剩餘:",'A'+j);
if(status==0)
{
avahead=available1=available2=(struct available*)malloc(avalen);
workhead=work1=work2=(struct available*)malloc(avalen);
available1->next=available2->next=NULL;
work1->next=work2->next=NULL;
scanf("%d",&available1->value);
work1->value=available1->value;
status++;
}
else
{
available2=(struct available*)malloc(avalen);
work2=(struct available*)malloc(avalen);
scanf("%d,%d",&available2->value);
work2->value=available2->value;
if(status==1)
{
avahead->next=available2;
workhead->next=work2;
status++;
}
available1->next=available2;
available1=available2;
work1->next=work2;
work1=work2;
}
}
available2->next=NULL;
work2->next=NULL;
status=0;
alloctemp=allochead;
maxtemp=maxhead;
for(i=0;i<row;i++)
for (j=0;j<colum;j++)
{
if(status==0)
{
needhead=need1=need2=(struct need*)malloc(needlen);
need1->next=need2->next=NULL;
need1->value=maxtemp->value-alloctemp->value;
status++;
}
else
{
need2=(struct need *)malloc(needlen);
need2->value=(maxtemp->value)-(alloctemp->value);
if(status==1)
{
needhead->next=need2;
status++;
}
need1->next=need2;
need1=need2;
}
maxtemp=maxtemp->next;
alloctemp=alloctemp->next;
}
need2->next=NULL;
status=0;
for(i=0;i<row;i++)
{
if(status==0)
{
finihead=finish1=finish2=(struct finish*)malloc(finilen);
finish1->next=finish2->next=NULL;
finish1->stat=0;
status++;
}
else
{
finish2=(struct finish*)malloc(finilen);
finish2->stat=0;
if(status==1)
{
finihead->next=finish2;
status++;
}
finish1->next=finish2;
finish1=finish2;
}
}
finish2->next=NULL; /*Initialization compleated*/
status=0;
processtest=0;
for(temp=0;temp<row;temp++)
{
alloctemp=allochead;
needtemp=needhead;
finishtemp=finihead;
worktemp=workhead;
for(i=0;i<row;i++)
{
worktemp1=worktemp;
if(finishtemp->stat==0)
{
for(j=0;j<colum;j++,needtemp=needtemp->next,worktemp=worktemp->next)
if(needtemp->value<=worktemp->value)
processtest++;
if(processtest==colum)
{
for(j=0;j<colum;j++)
{
worktemp1->value+=alloctemp->value;
worktemp1=worktemp1->next;
alloctemp=alloctemp->next;
}
if(status==0)
{
pathhead=path1=path2=(struct path*)malloc(pathlen);
path1->next=path2->next=NULL;
path1->value=i;
status++;
}
else
{
path2=(struct path*)malloc(pathlen);
path2->value=i;
if(status==1)
{
pathhead->next=path2;
status++;
}
path1->next=path2;
path1=path2;
}
finishtemp->stat=1;
}
else
{
for(t=0;t<colum;t++)
alloctemp=alloctemp->next;
finishtemp->stat=0;
}
}
else
for(t=0;t<colum;t++)
{
needtemp=needtemp->next;
alloctemp=alloctemp->next;
}
processtest=0;
worktemp=workhead;
finishtemp=finishtemp->next;
}
}
path2->next=NULL;
finishtemp=finihead;
for(temp=0;temp<row;temp++)
{
if(finishtemp->stat==0)
{
printf("\n系統處於非安全狀態!\n");
exit(0);
}
finishtemp=finishtemp->next;
}
printf("\n系統處於安全狀態.\n");
printf("\n安全序列為: \n");
do
{
printf("p%d ",pathhead->value);
}
while(pathhead=pathhead->next);
printf("\n");
return 0;
}