java.lang.NumberFormatException Fix? Strings in Integers

Question

*homework

Using the command line command

java class -add 7 8

will add all integers after "-add". But for this program, I am also suppose to print "Argument type mismatch" if anything after "-add" is not an integer.

I have this method to search a String for integers. If it finds something other than an integer, it returns false.

public static boolean isInteger(String s) {
    try {
        int num = Integer.parseInt(s);
        return true;

    } catch (Exception e) {}
    return false;
}

Now in this method I try to use the isInteger method to look at a String in the first for loop.

private static void add(String[] args) {
    for (int j = 1; j < args.length; j++) {
        if (isInteger(args[j]) == false)
        System.out.println("Argument type mismatch");
    }
    if (args.length == 1)
        System.out.println("Argument count mismatch");
    else {
        int result = 0;
        for (int i = 1; i < args.length; i++) {
            result += Integer.parseInt(args[i]);
        }
        System.out.println(result);
    }  
}

Running the first for loop by itself produces "Argument type mismatch" perfectly fine. But when I run the whole method and type the command

java class -add cat

it produces "Argument type mismatch" followed by a java.lang.NumberFormatException: For input string: "cat" error at the line result += Integer.parseInt(args[i]);

How can I fix this error?

Adding return; after System.out.println("Argument type mismatch"); should solve it. — Voicu
– Voicu, Commented Oct 7, 2014 at 1:11
This is because you are passing 2 arguments that is -add and cat to the class. The Integer.parseInt in else throws this error. — StackFlowed
– StackFlowed, Commented Oct 7, 2014 at 1:12

Bohemian · Accepted Answer · 2014-10-07 01:26:24Z

4

You get the error because you don't terminate exection once you find there's a problem with the input.

Add a return; statement to the code the executes when there's a problem:

for (int j = 1; j < args.length; j++) {
    if (!isInteger(args[j])) {
        System.out.println("Argument type mismatch");
        return; // Added this line
    }
}

Also note the slight change in the if condition. It's better coding style to not compare with boolean literals, ie change:

if (isInteger(args[j]) == false)

to

if (!isInteger(args[j]))

The reason this is good coding practice is it prevents this happening:

// Oops! An assignment that compiles and executes as "true"
if (someBooleanVariable = true)

when you meant to code a comparison:

if (someBooleanVariable == true)

The visual difference is subtle, and can lead to frustrating bugs if you don't visually pick it up, but there is no chance of this problem if you code:

if (someBooleanVariable)

edited Oct 7, 2014 at 1:26

answered Oct 7, 2014 at 1:12

Bohemian♦

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5 Comments

Jerry Over a year ago

Thanks for the tip on the if condition. Better to stop coding like this right when I'm in the learning process.

Bohemian Over a year ago

@Jerry yeah - see edit to answer for why it's not good practice

Jerry Over a year ago

Actually by doing this, the command java class -add 5 4 will no longer print anything when it should print 9.

Bohemian Over a year ago

@Jerry huh? Note the opening brace after the if and the close brace after the return. Have you got those in?

Jerry Over a year ago

I did not nice catch.

Simon Arsenault · Accepted Answer · 2014-10-07 01:17:51Z

2

It prints Argument type mismatch, but you do not end your program. You can add a return statement as @Bohemian suggested or you can use a if structure like below. I personally don't like return statement in a middle of a function. Choose a solution that fits you :)

Try something like this:

private static void add(String[] args) 
{
    if (args.length <= 1) // Validate number of arguments
    {
        System.out.println("Argument count mismatch");
    }
    else // Number of arguments valid
    {
        // Validate that we received only numbers
        boolean allIntegerValid = true;

        for (int j = 1; j < args.length; j++) 
        {
            allIntegerValid = allIntegerValid && isInteger(args[j]);
        }

        if (allIntegerValid) // We received only numbers
        {
            int result = 0;
            for (int i = 1; i < args.length; i++) 
            {
                result += Integer.parseInt(args[i]);
            }
            System.out.println(result);
        }
        else // At least one argument was not a number
        {
            System.out.println("Argument type mismatch");
        }
    } 
}

edited Oct 7, 2014 at 1:17

answered Oct 7, 2014 at 1:12

Simon Arsenault

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1 Comment

Kick Buttowski Over a year ago

I +1 you, but it would be nice if you do not give the code answer away :)

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