Golang Simple Implementation Banker algorithm

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This week the course design, thought will be very fun, the experiment book sent down, incredibly just realized banker algorithm, disappointed ... Used a small half a day (forgive me on the operating system incredibly even banker algorithm is what forget, can only go back to read a book again), with Golang simple implementation of the banker algorithm, is the code:

 PackageMainImport("FMT"    "Log"    "OS")varAvailable []int    //can use resource vectorsvarMax []int        //maximum demand matrixvarAllocation []int //Assigned matrixvarNeed []int       //Demand MatrixvarRequest []int      //Request Vectorvarwork []int         //working vectorvarFinish []BOOL      //finish VectorvarPidint            //Request Resource Process numbervarserial []int       //For storing secure sequencesvarwork []int//InitializefuncInit () {//Number of resources is 10,5,7    //Number of resources is 10,5,7}//Banker algorithmfuncBank () { forI: =0; I <Len(Request); i++ {ifRequest[i] <= Need[pid][i] {Continue}Else{log. Fatalln ("The required resources exceed the declared maximum value")        }    } forI: =0; I <Len(Request); i++ {ifRequest[i] <= Available[i] {Continue}Else{log. Fatalln ("There is not enough resources to wait")        }    } forI: =0; I <Len(Available); i++ {Available[i] = Available[i]-request[i]} forI: =0; I <Len(Allocation[0]); i++ {Allocation[pid][i] = Allocation[pid][i] + request[i]} forI: =0; I <Len(Need[0]); i++ {Need[pid][i] = Need[pid][i]-request[i]}ifSafe () {fmt. Println ("Complete Assignment")    }Else{FMT. Println ("The system is in an insecure state, waiting")} FMT. Println ("Max:", Max) fmt. Println ("Allocation:", Allocation) fmt. Println ("Need:", need) fmt. Println ("Available:", Available) fmt. Println ("Work:", work) FMT. Println ("Safe sequence:", serial)}//Security checkfuncSafe ()BOOL{work = Available forI: =0; I <Len(Max); i++ {Finish =Append(Finish,false)} Temp: =0Step2: forI: = temp; I <Len(Finish); i++ {ifFinish[i] = =false{ forJ: =0; J <Len(work); J + + {ifNEED[I][J] <= Work[j] {ifj = =Len(Work)-1{serial =Append(Serial, i) Work =Append(Work and work) forJ: =0; J <Len(work); J + + {Work[j] = Work[j] + allocation[i][j]} Fini Sh[i] =true                        ifi = =Len(Finish)-1{temp =0}Else{temp = i +1}GotoSTEP2}}Else{ Break}            }        }Else{Continue}ifi = =Len(Finish)-1{return false}    }return true}funcMain () {lable:varSelintFmt. Println ("******************** banker algorithm ********************") FMT. Println () fmt. Println ("\T1, test data One") FMT. Println ("\T2, test Data II") FMT. Println ("\T3, test data three") FMT. Println ("\t4, Exit") FMT. Println () fmt. Println ("************************************************") FMT. Print ("Please make your choice:") FMT. SCANLN (&sel)SwitchSEL {//The total number of resources is 10,5,7     Case1://test data 1Available = []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     Case2://test Data 2Available = []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     Case3://test Data 3Available = []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     Case4: OS. Exit(0)default: FMT. Println ("No this option")    }//Execution banker algorithmBank ()Gotolable}

Very humble, only tested three sets of data, if found a bug, look treatise, thank you!