JavaScript: What's the difference between .push(+var) and .push(var)

Question

I am splitting up a number into individual digits and placing them in an array:

var number = 123456789;                           //number I want to split
var output = [];                                  //empty output array
var sortedNumber = number.toString();             //converting number to string
var len = sortedNumber.length;

for (var i = 0; i < len; i += 1) {
    output.push(sortedNumber.charAt(i));          //Digits pushed into empty output array
}

console.log(output);

When I console.log the array, the array outputs the digits as strings, rather than actual numbers:

["1", "2", "3", "4", "5", "6", "7", "8", "9"]

However, when I adjust my code to:

output.push(+sortedNumber.charAt(i));

.. the array outputs the digits as actual numbers, which is what I want.

[1, 2, 3, 4, 5, 6, 7, 8, 9]

Why is this the case? What does the '+' do that means the array is full of numbers rather than strings? I am new to programming so any explanation would be greatly appreciated. Thanks in advance.

`charAt` returns a character as a string. By putting `+` in front of it you force it to become an integer (if it contains an actual number) — Federico klez Culloca, Sep 27 '18 at 09:43
@FedericoklezCulloca — There is no integer type in JS, it turns it into a number. — Quentin, Sep 27 '18 at 09:44
sidenotes: `0+sortedNumber.charAt(i)` will result in string `01`, `-sortedNumber.charAt(i)` will be number `-1`, `0-sortedNumber.charAt(i)` will _also_ be number `-1` — Jeff, Sep 27 '18 at 09:47
Also, if the string *doesn't* contain a number, the output of `+var` would be `NaN` (not a number). Its type is a number, which is somewhat counter-intuitive. So, you could say that the string will be converted to a number...although that number could also be the not a number value. — VLAZ, Sep 27 '18 at 09:47
you cast primitive to a different type. You can even do " " + 5 = "5", eg.. — Drahoš Maďar, Sep 27 '18 at 09:50
@vlaz if the string were empty, would the value still be NaN? — Caleb Whittington, Sep 27 '18 at 09:53
No, it would be zero. Also a bit counter intuitively, I guess. — VLAZ, Sep 27 '18 at 09:56

JavaScript: What's the difference between .push(+var) and .push(var)

0 Answers0