CODEWITHSUNDEEP
javascript
Brian Tucker
Brian Tucker

Reputation: 310

Basic ternary to if/then conversion in Javascript

Would someone mind converting this line of ternary code to if/then statements. I understand ternary, but am not able to get syntax errors to go away when I convert to if/then. This is the only line of my homework I had to borrow and I'd like to make it into if/then so that I can comment on it and understand it better myself.

Original:

return n == null || isNaN(n) ? 0 : n;

My attempt:

return n == null || if(isNaN(n)){return 0;}else{return n;}

Upvotes: 1

Views: 659

Answers (6)

wheresrhys
wheresrhys

Reputation: 23500

The problem is that return must be the first statement on any line it appears (as per the ECMA specs).

The code is easier to translate if you first add parentheses ( n == null || (isNaN(n) ? 0 : n) gives different results and is not equivalent to javascript's default parsing of un-parenthesesed code)

return (n == null || isNaN(n)) ? 0 : n;

which is equivalent to 

if (n == null || isNaN(n)) {
    return 0;
} else {
    return n;
}

Upvotes: 0

Bergi
Bergi

Reputation: 664405

You will need to apply some grammatics here:

The other four answers present the correct source code, I don't want to repeat it.

Upvotes: 0

xdazz
xdazz

Reputation: 160833

if (n == null || isNaN(n)) {
  return 0;
} else {
  return n;
}

Upvotes: 1

nnnnnn
nnnnnn

Reputation: 150020

This is what I think it is doing:

if (n == null || isNan(n))
   return 0;
else
   return n;

The || operator has higher precedence than the ? operator.

Upvotes: 0

Petar Ivanov
Petar Ivanov

Reputation: 93030

if (n == null || isNaN(n))
    return 0;
else 
    return n;

Upvotes: 0

Paul
Paul

Reputation: 141839

You have to move the return and have a separate one for the if and the else:

if(n == null || isNaN(n)){
    return 0;
}else{
    return n;
}

Upvotes: 0

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