Why my JavaScript code is printing undefined with the exception throwed?

Question

The Problem here is that with the output coming JavaScript also says undefined. I have seen some codes on w3schools.com and developers.mozilla websites and tried some other sites also but didn't find any solution or any explanation about the problem that i encounter.

I want to throw an exception when 0 or negative value occur.

function isPositive(a) {

  try {
    if (a > 0) return "YES";

    if (a < 0) throw "Negative Value";

    if (a == 0) throw "Zero Value";
  } catch (error) {
    console.log(error);
  }
}


console.log(isPositive(0));

Output expected was "Zero Value"

Actual Output is

"Zero Value" 

undefined

Any suggestions why my code prints undefined along with the output? Thanks in Advance.

Where are you running this? Also, will you please post the rest of your code. — wuno
– wuno, Commented Mar 26, 2019 at 6:30
You don't return anything unless the result is positive. And why are you throwing in a try? — Carcigenicate
– Carcigenicate, Commented Mar 26, 2019 at 6:32
this is all code that i have. i am testing my code on jsbin.com. I have already tried without try and catch but it was not working. — Amit Khatri
– Amit Khatri, Commented Mar 26, 2019 at 6:46
You have two console.log in that code. What do you expect the second one to log? — deceze
– deceze ♦, Commented Mar 26, 2019 at 7:02

Maheer Ali · Accepted Answer · 2019-03-26 06:33:01Z

1

This is because if any error is thrown in try it doesnot stop execution of code. If you want to stop the function on error just remove that try-catch and simply return

function isPositive(a){
    if(a>0) return "YES"
    if(a<0) return "Negative Value";
    if(a == 0) return "Zero Value";
}
console.log(isPositive(0));

If you want to throw error put the the statements out of try

function isPositive(a){
    if(a>0) return "YES"
    if(a<0) throw ("Negative Value");
    if(a == 0) throw ("Zero Value");
}
console.log(isPositive(0));

answered Mar 26, 2019 at 6:33

Maheer Ali

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2 Comments

Amit Khatri Over a year ago

I have also tried to put the if statements out of the try and catch block but it still doesn't work and giving "Uncaught Zero Value (line 12)".

Maheer Ali Over a year ago

@Amit Khatri So whats problem it that. If you want "Zero Value" only then use the first snippet

Amit Khatri · Accepted Answer · 2019-03-26 07:11:41Z

0

function isPositive(a){
try{
if(a>0) return "YES";

   if(a<0) throw "Negative Value";

   if(a == 0) throw "Zero Value";
}catch(err)
{
return err;
}
}


console.log(isPositive(0));

I find the answer and this code works for me and giving the proper results.

answered Mar 26, 2019 at 7:11

Amit Khatri

299 bronze badges

Comments

leonid066 · Accepted Answer · 2022-01-02 19:50:30Z

0

You have console.log(isPositive(a)) calling the function and if it throws error it returns "undefined" (and you will have it printed into console). But if you specify that you return the error it will print the "Negative Error". Code bellow.

function isPositive(a) {
    try{
        if (a > 0) return "YES" 
        if (a  == 0) throw ('Zero Error');    
        if (a < 0) throw ('Negative Error');
    }
    catch(error){
        return(error);
    }
}

answered Jan 2, 2022 at 19:50

leonid066

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