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I have this function (going trough the Eloquent Javascript Tutorial chapter 3):

function absolute(number) {
 if (number < 0)
 return -number;
 else
 return number;
}

show(absolute(prompt("Pick a number", "")));

If I run it and enter -3 the output will be 3 as expectet but if I enter just 3 the output will be "3" (with double quotes). I can get around by changing

return number;

to return Number(number);

but why is that necessary? What am I missing?

asked Feb 27, 2012 at 19:42
2
  • @mgraph: Try parseInt("042") Commented Feb 27, 2012 at 19:44
  • @NickBeranek: It's octal. You need to force it to parse as decimal by adding , 10 Commented Feb 27, 2012 at 19:49

4 Answers 4

2

prompt() always returns a string, but when you enter a negative number, it is handed to the -number call and implicitly converted to a Number. That doesn't happen if you pass it a positive, and the value received by prompt() is returned directly.

You can, as you discovered, cast it with Number(), or you can use parseInt(number, 10), or you could do -(-number) to flip it negative, then positive again, or more obviously as pointed out in comments, +number. (Don't do --number, which will cast it to a Number then decrement it)

answered Feb 27, 2012 at 19:45
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1 Comment

@SLaks Indeed, Overcomplication is my middle name.
1

Javascript is not strongly typed.

number comes from the prompt() function, which returns a string.
Since you aren't doing anything to change its type, it remains a string.

-number implicitly converts and returns an actual number.

answered Feb 27, 2012 at 19:43

Comments

1

If you have a string that needs to be converted to a number, please do the following:

var numString = '3';
var num = parseInt(numString);
console.log(num); // 3
answered Feb 27, 2012 at 19:45

Comments

0

JavaScript performs automatic conversion between types. Your incoming "number" is most likely string (you can verify by showing result of typeof(number)).

- does not take "string" as argument, so it will be converted to number first and than negated. You can get the same behavior using unary +: typeof(+ "3") is number when typeof("3") is string.

Same happens for binary - - will convert operands to number. + is more fun as it work with both strings "1"+"2" is "12", but 1+2 is 3.

answered Feb 27, 2012 at 19:51

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