程序要求输入double num 1和double num 2的用户输入 如果有异常,我希望它再次询问num 1和num 2的输入
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
double num1, num2;
int error = 0;
int text;
System.out.print("Enter 4 ");
text = sc.nextInt();
do{
try{
if(text == 4){
System.out.print("Enter number 1: ");
num1 = sc.nextDouble();
System.out.print("Enter number 2: ");
num2 = sc.nextDouble();
double quotient = num1/num2;
System.out.println("The Quotient of "+num1 + "/" +num2+ " = "+quotient);
}
}catch(Exception ex){
System.out.println("You've entered wrong input");
error = 1;
}
}while(error == 1);
}
然后当我尝试代码时,如果它会通过在num1或num 2中输入字符串来捕获异常,那么我有这个无限循环:
Enter number 1: You've entered wrong input
Enter number 1: You've entered wrong input
Enter number 1: You've entered wrong input
Enter number 1: You've entered wrong input
Enter number 1: You've entered wrong input
答案 0 :(得分:3)
您需要重置循环内的error
变量
do {
error = 0;
//...
} while(error == 1);
答案 1 :(得分:1)
它在C#中但相对类似:)
public class Program
{
private static double ReadUserInput (string message)
{
// This is a double
// The '?' makes it nullable which is easier to work with
double? input = null;
do
{
// Write message out
Console.Write(message);
// Read answer
var inputString = Console.ReadLine();
// Temp variable for the number
double outputNumber = 0;
// Try parse the number
if (double.TryParse(inputString, out outputNumber))
{
// The number was parsable as a double so lets set the input variable
input = outputNumber;
}
else
{
// Tell the user the number was invalid
Console.WriteLine("Sorry bud, but '" + inputString + "' is not a valid double");
}
}
while (input == null); // Keep running until the input variable is actually set by the above
// Return the output
return (double)input;
}
public static void Main()
{
// Read a number
var num1 = ReadUserInput("Enter number 1:");
// Read another number
var num2 = ReadUserInput("Enter number 2:");
// Show the calculation
Console.WriteLine("Answer: " + (num1*num2));
}
}
对于实际代码(在JAVA中):
public class JavaFiddle
{
public static void main (String[] args)
{
// Read a number
Double num1 = ReadUserInput("Enter number 1:");
// Read another number
Double num2 = ReadUserInput("Enter number 2:");
// Show the calculation
System.out.println("Answer: " + (num1*num2));
}
public static Double ReadUserInput (String message)
{
java.util.Scanner inputScanner = new java.util.Scanner(System.in);
Double input = null;
do
{
// Write message out
System.out.println(message);
// Read answer
String inputString = inputScanner.nextLine();
try
{
// Try parse the number
input = Double.parseDouble(inputString);
}
catch (NumberFormatException e)
{
// Tell the user the number was invalid
System.out.println("Sorry bud, but '" + inputString + "' is not a valid double");
}
}
while (input == null); // Keep running until the input variable is actually set by the above
// Return the output
return input;
}
}
答案 2 :(得分:1)
没有必要使用异常处理。只需使用Scanner.hasNextDouble()
方法查明实际用户输入是否为double,否则继续循环。
package com.company;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
double num1, num2;
num1 = readDouble(1, sc);
num2 = readDouble(2, sc);
double quotient = num1/num2;
System.out.println("The Quotient of " + num1 + "/" + num2 + " = " + quotient);
}
private static double readDouble(int i, Scanner sc) {
while (true) {
System.out.print("Enter number " + i + ": ");
if (!sc.hasNextDouble()) {
System.out.println("You've entered wrong input");
sc.next();
continue;
}
break;
}
return sc.nextDouble();
}
}
答案 3 :(得分:0)
您可能想测试是否没有错误:
}while(error != 1);
或
}while(error == 0);
答案 4 :(得分:0)
如果输入无效,您将需要一个调用自身的输入方法。
double getInput(Scanner sc) {
try {
double num = sc.nextDouble();
return num;
} catch(Exception ex) {
System.out.println("You've entered wrong input");
return getInput(sc);
}
}
并在您的其他方法中调用此方法两次。
答案 5 :(得分:0)
它可能看起来很难看,但这是一种方法
do
{
if(...)
{
boolean successReading = false;
while(!successReading)
{
try
{
System.out.print("Enter number 1: ");
num1 = sc.nextDouble();
System.out.print("Enter number 2: ");
num2 = sc.nextDouble();
successReading = true;
double product = num1*num2;
}
catch(Exception e)
{
successReading = false;
}
}
}
}while(...)
答案 6 :(得分:0)
您需要在sc.next();
阻止内添加catch
。
nextDouble
方法不清除缓冲区。因此,下次调用它时会出现相同的错误,因为旧输入仍在缓冲区中。
您还需要在循环开始时重置error
标志。
答案 7 :(得分:0)
您必须将sc.next();
放入陷阱中,这样才能清除扫描仪变量并要求输入