我正在尝试创建一个简单的基于文本的操作系统，但我无法弄清楚为什么我的代码在计算器类完成后不允许我输入命令。它应该继续执行代码，直到我输入“off”但事实并非如此。 Eclipse说它正在运行，但我什么都做不了。有人可以帮帮我吗？
这是我的两个班级：
public class Calculator extends Start{
public static void calStrt() {
System.out.print("\nEnter operator you wish to use: ");
StringInput = scan.nextLine();
if (StringInput.equals("+")) {
add();
} else if (StringInput.equals("-")) {
sub();
} else if (StringInput.equals("*")) {
mul();
} else if (StringInput.equals("/")) {
div();
} else {
System.out.println("\nSyntax error: Operator not recognized");
System.out.println("Please try again");
calStrt();
}
}
public static void add() {
System.out.print("\nEnter first number: ");
intInput = scan.nextInt();
int intVar1 = intInput;
System.out.print("\nEnter second number: ");
intInput = scan.nextInt();
int intVar2 = intInput;
System.out.println("\nAnswer: " + (intVar1 + intVar2));
}
public static void sub() {
System.out.print("\nEnter first number: ");
intInput = scan.nextInt();
int intVar1 = intInput;
System.out.print("\nEnter second number: ");
intInput = scan.nextInt();
int intVar2 = intInput;
System.out.println("\nAnswer: " + (intVar1 - intVar2));
}
public static void mul() {
System.out.print("\nEnter first number: ");
intInput = scan.nextInt();
int intVar1 = intInput;
System.out.print("\nEnter second number: ");
intInput = scan.nextInt();
int intVar2 = intInput;
System.out.println("\nAnswer: " + (intVar1 * intVar2));
}
public static void div() {
System.out.print("\nEnter first number: ");
intInput = scan.nextInt();
int intVar1 = intInput;
System.out.print("\nEnter second number: ");
intInput = scan.nextInt();
int intVar2 = intInput;
System.out.println("\nAnswer: " + (intVar1 / intVar2));
}
}
import java.util.Scanner;
class Start {
static Scanner scan = new Scanner(System.in);
static String StringInput;
static int intInput;
public static void main(String[] args) {
System.out.println("\nWelcome to RobOS");
passLoop: while (true) {
System.out.print("\nPlease enter password: ");
StringInput = scan.nextLine();
if (StringInput.equals("banana")) {
System.out.print("Logging in, please wait");
System.out.print(".");
System.out.print(".");
System.out.println(".");
System.out.println("\nWelcome User");
outerLoop: while (true) {
System.out.println("\nType \"help\" to see a list of programs");
StringInput = scan.nextLine();
innerLoop: while (true) {
if (StringInput.equalsIgnoreCase("cal")) {
Calculator.calStrt();
continue outerLoop;
} else if (StringInput.equalsIgnoreCase("guess")) {
GuessGame.guess();
continue outerLoop;
} else if (StringInput.equalsIgnoreCase("help")) {
System.out.println("\n\"cal\" uses the calculator");
System.out.println("\"guess\" plays guessing game");
System.out.println("\"help\" shows list of programs");
System.out.println("\"off\" turns RobOS off");
continue outerLoop;
} else if (StringInput.equalsIgnoreCase("off")){
break passLoop;
}
}
}
} else {
System.out.println("\nWrong password. Please try again");
continue passLoop;
}
}
}
}
答案 0 :(得分：2)
scan.nextInt()：Integer.parseInt(scan.nextLine());
希望这有效
答案 1 :(得分：1)
您的代码进入无限循环。当您致电StringInput = scan.nextLine()时，它第一次正常工作。我输入了cal，我可以运行一次计算器。问题是第二次scan.nextLine()被调用，它会自动返回一个空字符串“”作为StringInput的值。 if/else中的while(true)语句集合无法处理此问题，因此它只是永远循环。
更深层的理由是你打电话给scan.nextInt()读取数字，但问题是当你读到计算器操作的第二个数字时，{\ n}上还有一个“\ n” }。因此，当您循环并再次调用System.in时，它不会提示您输入任何内容，因为它只读取仍然位于scan.nextLine()的“\ n”然后将您发送到无限循环。