parseInt函数在一个阵列

Question

import java.util.Scanner; 

public class NumAverage { 

    public static void main (String [] args) 
    { 
     //int a,b; 
     int[] numbers = new int [10]; 

     Scanner numreader = new Scanner(System.in); 

     try 
     { 
      System.out.println("Enter 10 num"); 
      for (int i = 0; i < numbers.length; i++) 
       numbers[i] = numreader.nextInt(); 
       //numbers[i] = Integer.parseInt(s) 
      //for (int i = 0; i < numbers.length; i++) 
      //{ 
       //sum += numbers[i]; 
      //} 
     } 
     catch(NumberFormatException numfo) 
     { 
      System.out.println(numfo.getMessage() + "cannot be converted to integer"); 
     } 
    } 
} 

我怎样才能得到我已经被从键盘inputed到数组是parseInt函数，如果用户尝试输入一个字符串，而不是整数

Answer 1

Integer.parse的数字（字符串如果s不是整数，则抛出一个NumberFormatException异常。 所以围绕字符串到int的转换与一个try-catch子句

Answer 2

尝试：

import java.util.Scanner; 

public class NumAverage { 

public static void main (String [] args) 
{ 
    //int a,b; 
    int[] numbers = new int [10]; 

    Scanner numreader = new Scanner(System.in); 
    System.out.println("Enter 10 num"); 
    for (int i = 0; i < numbers.length; i++) { 
     // Without the try and catch you get the appropriate exception, 
     // NumberFormatException in this case 
     String str = numreader.next(); 
     numbers[i] = Integer.parseInt(str); 

     //try { 
     // numbers[i] = Integer.parseInt(numreader.next()); 
     //} catch(NumberFormatException e) { 
     // number[i] = 0; // or whatever you want 
     //} 
    } 
    } 
} 

Answer 3

这是我会为你做的情况下。

int[] numbers = new int[10]; 

Scanner numreader = new Scanner(System.in); 

System.out.println("Enter 10 numbers"); 

// Get User Input 
for (int i = 0; i < numbers.length; i++) { 
    try { 
     numbers[i] = Integer.parseInt((String) numreader.nextLine()); 
    } catch (NumberFormatException numFormatE) { 
     System.out.println(numFormatE.getMessage() + "cannot be converted to integer"); 
     i = i - 1; // Restart current iteration. 
    } 
} 
numreader.close(); 

// Loop through the number to get sum of the number. 
int sum = 0; 
for (int i = 0; i < numbers.length; i++) { 
    sum += numbers[i]; 
} 
System.out.println(sum); 

但是，如果你想总和在主回路结合起来，这应该工作以及：

int[] numbers = new int[10]; 
int sum = 0; 

Scanner numreader = new Scanner(System.in); 

System.out.println("Enter 10 numbers"); 

// Get User Input 
for (int i = 0; i < numbers.length; i++) { 
    try { 
     numbers[i] = Integer.parseInt((String) numreader.nextLine()); 
    } catch (NumberFormatException numFormatE) { 
     System.out.println(numFormatE.getMessage() + "cannot be converted to integer"); 
     i = i - 1; // Restart current iteration. 
    } 
    sum += numbers[i]; // Add the number in loop 
} 
System.out.println(sum); 
numreader.close(); 

更新：修正代码格式化