Java如何从try，catch和finally返回值？ [英] Java How do you return a value from try,catch, and finally?

问题描述

所以当我在try {}中做一个块的代码时，我尝试返回一个值，它出现了不返回值。这是我使用的代码代表我的问题。

  import org.w3c.dom.ranges.RangeException; 
 
 
 public class Pg257E5 
 {
 public static void main（String [] args）
 {
 
 try 
 {
 System.out.println（add（args））; 
} 
 catch（RangeException e）
 {
 e.printStackTrace（）; 
} 
 finally 
 {
 System.out.println（感谢使用程序kiddo！）; 
} 
 
} 
 public static double add（String [] values）//这里显示一个commpile错误，我没有一个返回值
 {
 try 
 {
 int length = values.length; 
 double arrayValues [] = new double [length]; 
 double sum = 0; （int i = 0; i< length; i ++）
 {
 arrayValues [i] = Double.parseDouble（values [i]）; 
 
 sum + = arrayValues [i]; 
} 
 
返还金额//这里有一个返回值。是因为如果一个异常发生在try中的代码，并且没有得到返回值？ 
} 
 catch（NumberFormatException e）
 {
 e.printStackTrace（）; 
} 
 catch（RangeException e）
 {
 throw e; 
} 
 finally 
 {
 System.out.println（感谢您使用该程序！）; //所以我需要返回一个double类型的返回值这里？ 
} 
 
} 
} 
  

我遇到的问题是在使用try和catch块时如何返回值？

解决方案

要返回使用 try / catch 的值可以使用临时变量，例如

  public static double add（String [] values）{
 double sum = 0.0; 
 try {
 int length = values.length; 
 double arrayValues [] = new double对于（int i = 0; i< length; i ++）{
 arrayValues [i] = Double.parseDouble（values [i]）; 
 sum + = arrayValues [ i]; 
} catch（NumberFormatException e）{
 e.printStackTrace（）; 
} catch（RangeException e）{
 throw e; 
} finally {
 System.out.println（谢谢你使用程序！）; 
} 
 return sum; 
} 
  pre> 
 
 
否则您需要返回在每个执行路径（try block或catch块）中，没有 throw 。
 
So when I do a code of blocks inside a "try{}", and I try to return a value, it comes out " no return values". This is the code I am using that represents my problem.
import org.w3c.dom.ranges.RangeException;


public class Pg257E5 
{
public static void main(String[]args)
{

    try
    {
        System.out.println(add(args));
    }
    catch(RangeException e)
    {
        e.printStackTrace();
    }
    finally
    {
        System.out.println("Thanks for using the program kiddo!");
    }

}
public static double add(String[] values) // shows a commpile error here that I don't have a return value
{
    try
    {
        int length = values.length;
        double arrayValues[] = new double[length];
        double sum =0;
        for(int i = 0; i<length; i++)
        {
            arrayValues[i] = Double.parseDouble(values[i]);
            sum += arrayValues[i];
        }

        return sum; // I do have a return value here. Is it because if the an exception occurs the codes in try stops and doesn't get to the return value?
    }
    catch(NumberFormatException e)
    {
        e.printStackTrace();
    }
    catch(RangeException e)
    {
        throw e;
    }
    finally
    {
        System.out.println("Thank you for using the program!");// so would I need to put a return value of type double here?
    }

}
}
Bascially, the question I'm having is "How do you return a value when you are using try and catch block?
解决方案
To return a value when using try/catch you can use a temporary variable, e.g.
public static double add(String[] values) {
    double sum = 0.0;
    try {
        int length = values.length;
        double arrayValues[] = new double[length];
        for(int i = 0; i<length; i++) {
        arrayValues[i] = Double.parseDouble(values[i]);
        sum += arrayValues[i];
    } catch(NumberFormatException e) {
        e.printStackTrace();
    } catch(RangeException e) {
        throw e;
    } finally {
        System.out.println("Thank you for using the program!");
    }
    return sum;
}
Else you need to have a return in every execution path (try block or catch block) that has no throw.

                        
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