java - 如何从 try、catch 和 finally 中返回一个值？

Question

所以当我在 try{} 中执行代码块，并尝试返回一个值时，它告诉我

no return values

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?
    }

}
}

我的问题是，当您使用 try 和 catch 时如何返回值？

Answer 1

要在使用 try/catch 时返回值，您可以使用临时变量，例如

public static double add(String[] values) {
    double sum = 0.0;
    try {
        int length = values.length;
        double arrayValues[] = new double[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;
}

否则，您需要在每个没有throw 的执行路径(try block 或catch block )中都有一个return。