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 mainimport (    "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10,5,7}//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 := 0step2:    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10,5,7    case 1:        //测试数据1        Available = []int{3, 3, 2}        Max = [][]int{{7, 5, 3}, {3, 2, 2}, {9, 0, 2}, {2, 2, 2}, {4, 3, 3}}        Allocation = [][]int{{0, 1, 0}, {2, 0, 0}, {3, 0, 2}, {2, 1, 1}, {0, 0, 2}}        Need = [][]int{{7, 4, 3}, {1, 2, 2}, {6, 0, 0}, {0, 1, 1}, {4, 3, 1}}        Request = []int{1, 0, 2}        pid = 1    case 2:        //测试数据2        Available = []int{2, 3, 0}        Max = [][]int{{7, 5, 3}, {3, 2, 2}, {9, 0, 2}, {2, 2, 2}, {4, 3, 3}}        Allocation = [][]int{{0, 1, 0}, {3, 0, 2}, {3, 0, 2}, {2, 1, 1}, {0, 0, 2}}        Need = [][]int{{7, 4, 3}, {0, 2, 0}, {6, 0, 0}, {0, 1, 1}, {4, 3, 1}}        Request = []int{3, 3, 0}        pid = 4    case 3:        //测试数据3        Available = []int{2, 3, 0}        Max = [][]int{{7, 5, 3}, {3, 2, 2}, {9, 0, 2}, {2, 2, 2}, {4, 3, 3}}        Allocation = [][]int{{0, 1, 0}, {3, 0, 2}, {3, 0, 2}, {2, 1, 1}, {0, 0, 2}}        Need = [][]int{{7, 4, 3}, {0, 2, 0}, {6, 0, 0}, {0, 1, 1}, {4, 3, 1}}        Request = []int{0, 2, 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！