[Operating system] bankers algorithm

Banker's algorithm

Blog address: https://www.iamzlt.com/?p=87

(A) the purpose and requirements

Bankers avoid deadlock algorithm is an algorithm by Dijkstra most representative design. In this study, the simulation algorithm requested to prepare a banker in a high level language. We may have a more profound understanding of deadlock prevention and bankers algorithm by this experiment.

(B) experimental content

1 , in the data structure

Vector comprising resources (Availiable), maximum demand matrix (Max), the assignment matrix (Allocation), demand matrix (Need)

2 , algorithm design security

Set work indicates that the system provides vector Work process continues to run the number of available resources, Finish indicate whether the system has enough resources allocated to the process

(C) experimental environment

1. PC
2. VC ++

[Run] Code

#include<iostream>

#include<bits/stdc++.h>

#define MAXN 100

#define NAME_MAX

using namespace std;

const int PNumber=5;

const int SNumber=3;

char Source[PNumber];

int Request[SNumber];

int Available[SNumber]={3,3,2};

int MAX[PNumber][SNumber]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}};

int Allocation[PNumber][SNumber]={{0,1,0},{2,0,0},{3,0,2},{2,1,1},{0,0,2}};

int Need[PNumber][SNumber]={{7,4,3},{1,2,2},{6,0,0},{0,1,1},{4,3,1}};

int OP[PNumber][SNumber];

//int Over[PNumber];




bool judgement(int n){

    int flag=0;

    for(int j=0; j<SNumber; j++)

        if(Need[n][j]==0) flag++;

    if(flag==SNumber)return false;

    return true;

}




void initialize(){

    int i;

    for(i=0; i<PNumber; i++)

        if(!judgement(i)){

            for(int j=0; j<SNumber; j++) Available[j]+=MAX[i][j];

            cout<<"进程P"<<i+1<<"所需的资源全部满足，此进程运行完毕！"<<endl;

        }

    cout<<"此刻进程中存在的进程："<<endl;

    for(i=0; i<PNumber; i++)

        if(judgement(i))    cout<<"P"<<i+1<<'\t';







    cout<<endl<<"此刻系统可利用的资源（单位：个）："<<endl;

    for(i=0; i<SNumber; i++)

        printf("%c\t",'A'+i);

    cout<<endl;

    for(i=0; i<SNumber; i++)

        cout<<Available[i]<<'\t';

    cout<<endl<<"此刻各进程已占有资源如下（单位：个）："<<endl;

    for(i=0; i<SNumber; i++)

        printf("\t%c",'A'+i);

    cout<<endl;

    for(i=0; i<PNumber; i++)

        if(judgement(i)){

            cout<<"P"<<i+1<<'\t';

            for(int j=0; j<SNumber; j++)

                cout<<MAX[i][j]-Need[i][j]<<'\t';

            cout<<endl;

        }

    cout<<"各进程运行完毕还需各资源如下（单位：个）："<<endl;

    for(i=0; i<SNumber; i++)

        printf("\t%c",'A'+i);

    cout<<endl;

    for(i=0; i<PNumber; i++)

        if(judgement(i)){

            cout<<"P"<<i+1<<'\t';

            for(int j=0; j<SNumber; j++)

                cout<<Need[i][j]<<'\t';

            cout<<endl;

        }

}

void systemstatus(){




}

int security(int n){

    int Work[SNumber],i;

    bool Finish[PNumber];

    for(i=0; i<PNumber; i++)

        Finish[i]=false;

    for(i=0; i<SNumber; i++)

        Work[i]=Available[i];

    for(i=0; i<PNumber; i++){

        int flag=1;

        for(int j=0; j<SNumber; j++)

            if(!(Need[i][j]<=Work[j])) flag=0;

        if(!Finish[i]&&flag){

            for(int j=0; j<SNumber; j++)

                Work[j]+=Allocation[i][j];

            Finish[i]=true;

            //cout<<i<<endl;

            i=-1;

            continue;

        }

    }

    int flag=1;

    for(i=0; i<PNumber; i++)

        if(!Finish[i]) flag=0;

    if(flag==1){

            //

    }else{

        cout<<"不能满足申请，此进程挂起，原因为："<<endl;

        cout<<"若满足申请，系统将进入不安全状态，可能导致死锁！"<<endl;

        return 0;

    }

    return 1;

}




void banker(int n){

    for(int i=0; i<SNumber; i++)

        if(!(Request[i]<=Need[n][i])){

            cout<<"不能满足申请，此进程挂起，原因为："<<endl;

            cout<<"申请的资源中某种资源大于其声明的需求量！"<<endl;

            return;

        }

    for(int i=0; i<SNumber; i++)

        if(!(Request[i]<=Available[i])){

            cout<<"不能满足申请，此进程挂起，原因为："<<endl;

            cout<<"申请的资源大于系统可提供的资源量，P"<<n+1<<"需等待"<<endl;

            return;

        }

    for(int i=0; i<SNumber; i++){

        Available[i]-=Request[i];

        Allocation[n][i]+=Request[i];

        Need[n][i]-=Request[i];

    }

    if(security(n)==1){

        cout<<"可以满足申请，";

        initialize();

    }else{

        for(int i=0; i<SNumber; i++){

            Available[i]+=Request[i];

            Allocation[n][i]-=Request[i];

            Need[n][i]+=Request[i];

        }

    }

}




int main(){

    initialize();

    int n,i;

    while(1){

        cout<<"输入发出请求的进程（输入‘0’退出系统）：";

        cin>>n;

        if(n==0) break;

        cout<<"此进程申请个资源数目："<<endl;

        for(i=0; i<SNumber; i++){

            printf("%c",'A'+i);

            cout<<"资源：";

            cin>>Request[i];

        }

        banker(n-1);

    }

}

【operation result】

【Result analysis】

First: P4 resource request, P4 request vector (0,1,0), to inspect the banker's algorithm

1. Request(0,1,0)<=Need(0,1,1,)
2. Request(0,1,0)<=Availabele(3,2,2)
3. P4 system first assumed to allocate resources and modify values
4. Use security algorithm checks whether this system is safe

 

Similarly the second, the third can be determined.

In the last, the resource request P2, P4 request vector (1,2,2), to meet the application and the resources needed to meet all of P2, finished running will release its share of resources, system resources are available at this time {5,2,2}.