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?
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)
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.
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
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.
parseInt("042"), 10