golang简单实现银行家算法

这周课程设计，以为会很好玩，实验书发下来，居然只是实现银行家算法，失望了……用了小半天时间（原谅我上过操作系统居然连银行家算法是什么都忘了，只能回去看书重来），用golang简单实现了银行家算法，一下是代码：

package main

import (
    "fmt"
    "log"
    "os"
)

var Available []int    //可以用资源向量
var Max [][]int        //最大需求矩阵
var Allocation [][]int //已分配矩阵
var Need [][]int       //需求矩阵
var Request []int      //请求向量
var Work []int         //工作向量
var Finish []bool      //Finish向量
var pid int            //请求资源进程号
var serial []int       //用于存储安全序列
var work [][]int

//初始化
func init() {

    //资源数量为10,5,7

    //资源数量为10,7

}

//银行家算法
func Bank() {
    for i := 0; i < len(Request); i++ {
        if Request[i] <= Need[pid][i] {
            continue
        } else {
            log.Fatalln("所需资源超过宣布最大值")
        }
    }

    for i := 0; i < len(Request); i++ {
        if Request[i] <= Available[i] {
            continue
        } else {
            log.Fatalln("尚无足够资源，需等待")
        }
    }

    for i := 0; i < len(Available); i++ {
        Available[i] = Available[i] - Request[i]
    }

    for i := 0; i < len(Allocation[0]); i++ {
        Allocation[pid][i] = Allocation[pid][i] + Request[i]
    }

    for i := 0; i < len(Need[0]); i++ {
        Need[pid][i] = Need[pid][i] - Request[i]
    }

    if Safe() {
        fmt.Println("完成分配")
    } else {
        fmt.Println("系统处于不安全状态，等待")
    }

    fmt.Println("Max:",Max)
    fmt.Println("Allocation:",Allocation)
    fmt.Println("Need:",Need)
    fmt.Println("Available:",Available)
    fmt.Println("Work:",work)
    fmt.Println("安全序列：",serial)
}

//安全性检查
func Safe() bool {
    Work = Available
    for i := 0; i < len(Max); i++ {
        Finish = append(Finish,false)
    }

    temp := 0
step2:
    for i := temp; i < len(Finish); i++ {
        if Finish[i] == false {
            for j := 0; j < len(Work); j++ {
                if Need[i][j] <= Work[j] {
                    if j == len(Work)-1 {
                        serial = append(serial,i)
                        work = append(work,Work)

                        for j := 0; j < len(Work); j++ {
                            Work[j] = Work[j] + Allocation[i][j]
                        }
                        Finish[i] = true
                        if i == len(Finish)-1 {
                            temp = 0
                        } else {
                            temp = i + 1
                        }
                        goto step2
                    }
                } else {
                    break
                }
            }
        } else {
            continue
        }

        if i == len(Finish)-1 {
            return false
        }
    }

    return true
}

func main() {
lable:
    var sel int
    fmt.Println("********************银行家算法********************")
    fmt.Println()
    fmt.Println("\t1、测试数据一")
    fmt.Println("\t2、测试数据二")
    fmt.Println("\t3、测试数据三")
    fmt.Println("\t4、退出")
    fmt.Println()
    fmt.Println("************************************************")
    fmt.Print("请做出你的选择：")
    fmt.Scanln(&sel)
    switch sel {
    //资源总数量为10,7
    case 1:
        //测试数据1
        Available = []int{3, 3, 2}
        Max = [][]int{{7, 5, 3},{3, 2, 2},{9, 0,{2,{4, 3}}
        Allocation = [][]int{{0, 1, 0}, 1},{0, 2}}
        Need = [][]int{{7, 4,{1,{6, 1}}
        Request = []int{1, 2}
        pid = 1
    case 2:
        //测试数据2
        Available = []int{2, 0}
        Max = [][]int{{7, 1}}
        Request = []int{3, 0}
        pid = 4
    case 3:
        //测试数据3
        Available = []int{2, 1}}
        Request = []int{0, 0}
        pid = 0
    case 4:
        os.Exit(0)
    default:
        fmt.Println("没有此选项")
    }

    //执行银行家算法
    Bank()
    goto lable
}

很简陋，只测试了三组数据，如果发现bug，望斧正，谢谢！