Confusing JavaScript statement about string Concatenation

Question

I was developing a node.js site and I made a copy and paste error that resulted in the following line (simplified for this question):

var x = "hi" + + "mom"

It doesn't crash and x = NaN. Now that i have fixed this bug, I am curious what is going on here, since if I remove the space between the + signs I get an error (SyntaxError: invalid increment operand)

My Question is : Can some explain to me what is going on in the statement and how nothing (a space between the + signs) changes this from an error to a NaN?

PS. I am not sure if this should go here or programers.stackoverflow.com. Let me know if I posted on the wrong site.

My complete guess is the whitespace is interpreted as 0 or something? ---- scratch that. Its the result of "hi" + (+ "mom") I guess (+ "mom") is interpreted as positive "mom" and therefore NaN. — Drew R
– Drew R, Commented Dec 17, 2014 at 20:48
@Sirko Thank you for your quick response. The fact that it is coming back as NaN (or "hiNaN") is reasonable. I am just curious how is it getting to NaN instead of crashing. What javascript quark is this doing? — Dewseph
– Dewseph, Commented Dec 17, 2014 at 20:52
@AlexeyLebedev Thank you for answering, but could you explain your comment a little more please? — Dewseph
– Dewseph, Commented Dec 17, 2014 at 20:52
In a + + b, you have the addition operator and the unary + operator. More info here: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — Felix Kling
– Felix Kling, Commented Dec 17, 2014 at 20:58

Evan Davis · Accepted Answer · 2014-12-17 21:09:49Z

7

It's being interpreted like this:

var x = "hi" + (+"mom")

The prefix + tries to coerce the string to a number. Number('mom') is NaN, so +'mom' is also NaN.

edited Dec 17, 2014 at 21:09

answered Dec 17, 2014 at 20:50

Evan Davis

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1 Comment

Felix Kling Over a year ago

"parseFloat('mom') is NaN, so +'mom' is also NaN." Rather "Number('mom') is NaN, so +'mom' is also NaN." Because parseFloat('5foo') is 5, but +'5foo' is still NaN.

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