JavaScript Exception not working

Question

I'm trying to figure out why my code does not throw and display the error message (page is just blank) after I call it with the following statement:

document.write(add(10,wrong_input));

program.js

var add = function (a,b){
    if(typeof a !== 'number' || typeof b !== 'number'){
        throw{
            name: 'TypeError',
            message: 'add needs numbers'
        } catch(e){
            document.writeln(e.name + ': ' + e.message);
        }
    }
    return a + b;
}

program.html

<html>
    <body>
    <pre><script src="program.js"></script></pre>
    <div></div>
    </body>
</html>

Where are you looking for the error message?

Quentin
– Quentin

2013-03-26 20:23:09 +00:00
Commented Mar 26, 2013 at 20:23 — Quentin
– Quentin, Commented Mar 26, 2013 at 20:23
You're mixing throw and try.

bfavaretto
– bfavaretto

2013-03-26 20:23:28 +00:00
Commented Mar 26, 2013 at 20:23 — bfavaretto
– bfavaretto, Commented Mar 26, 2013 at 20:23
I'm looking for it to be display in the browser

Anthony
– Anthony

2013-03-26 20:23:42 +00:00
Commented Mar 26, 2013 at 20:23 — Anthony
– Anthony, Commented Mar 26, 2013 at 20:23

bfavaretto · Accepted Answer · 2013-03-26 20:25:36Z

6

The throw statement doesn't have a catch clause, try does. You should throw and catch separately. For example:

var add = function (a,b){
    if(typeof a !== 'number' || typeof b !== 'number'){
        throw{
            name: 'TypeError',
            message: 'add needs numbers'
        }
    }
    return a + b;
}

try {
    add('foo', 1);
} catch(ex) {
    alert(ex.message);
}

Note that I replaced document.writeln with alert, because the former will overwrite the whole document if it runs after page load. If you want something better looking, manipulate the DOM directly (by changing some element's innerHTML, appending a node, etc).

answered Mar 26, 2013 at 20:25

bfavaretto

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Comments

Mathew Thompson · Accepted Answer · 2013-03-26 20:26:07Z

2

Your error handling code is slightly off, you can't throw an error and then try and write it out. You either do:

if(typeof a !== 'number' || typeof b !== 'number'){
    document.write("TypeError" + ": " + "add needs numbers");
}

Or simply just throw it:

if(typeof a !== 'number' || typeof b !== 'number'){
    throw {
        message: "add needs numbers",
        name: "TypeError"
    }
}

Then do your try catch in your function call. Personally though, I'd say stick with the first.

answered Mar 26, 2013 at 20:26

Mathew Thompson

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Comments

Bertrand · Accepted Answer · 2013-03-26 20:34:50Z

1

AS commented by bfaretto, You are mixing throw and try.

throw throws a exception that you define, but you are using it as a try..catch block. Here is how you could use throw and try..catch together.

var add = function (a,b){
    try {
       if(typeof a !== 'number' || typeof b !== 'number'){
            var n = {
                name: 'TypeError',
                message: 'add needs numbers'
            };
            throw n;
        }
        // throws an exception with a numeric value
    } catch (e) {
       console.log(e.name);
    }
}

answered Mar 26, 2013 at 20:34

Bertrand

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