[英]program finishes after exception is handled in Java
This is for homework. 这是为了功课。
I am trying to create a program that takes the average of ten input numbers. 我正在尝试创建一个平均需要输入10个数字的程序。 When the user enters a character that is not a number, the NumberFormatException exception is caught.
当用户输入非数字字符时,将捕获NumberFormatException异常。 The program was finishing after the exception was caught, so I changed it to use recursion to call the method again after the exception is caught, but now it prints multiple averages, some of which are not correct.
捕获到异常后程序正在完成,因此我将其更改为在捕获到异常后使用递归再次调用该方法,但是现在它会打印多个平均值,其中一些不正确。
How do I change the program so that it continues to ask for input after an exception has been caught instead of finishing? 如何更改程序,以便在捕获到异常而不是完成之后继续请求输入? I do not want the program to finish after catching the exception.
我不希望程序在捕获异常后完成。
import java.util.Scanner;
import java.util.Scanner;
public class PrintAverage {
int average;
public static void main(String args[]) {
System.out.println("You are going to enter ten numbers to find their average.");
getInput();
}
private static void getInput() {
String input;
int sum = 0;
int[] arrayOfIntegers = new int[10];
double average = 0;
Scanner scanner = new Scanner(System.in);
try {
for (int i = 0; i < arrayOfIntegers.length; i++) {
System.out.println("Enter the next number.");
input = scanner.nextLine();
arrayOfIntegers[i] = Integer.parseInt(input);
}
} catch (NumberFormatException exception) {
System.out.println("The last entry was not a valid number.");
getInput();
}
for (int k = 0; k < arrayOfIntegers.length; k++) {
sum = sum + arrayOfIntegers[k];
}
average = sum / arrayOfIntegers.length;
System.out.println("The average is " + average);
}
}
Your try/catch is not specific enough, so it catches the exception after asking all the numbers you want. 您的try / catch不够具体,因此在询问所有想要的数字后会捕获异常。
Localize the exception. 本地化异常。 Change this:
更改此:
try {
for (int i = 0; i < arrayOfIntegers.length; i++) {
System.out.println("Enter the next number.");
input = scanner.nextLine();
arrayOfIntegers[i] = Integer.parseInt(input);
}
} catch (NumberFormatException exception) {
System.out.println("The last entry was not a valid number.");
getInput();
}
to this: 对此:
for (int i = 0; i < arrayOfIntegers.length; i++) {
System.out.println("Enter the next number.");
input = scanner.nextLine();
try {
arrayOfIntegers[i] = Integer.parseInt(input);
} catch (NumberFormatException exception) {
System.out.println("The last entry was not a valid number.");
i--; //so you don't lose one of the 10.
}
}
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