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Home/ Questions/Q 7183837
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T18:04:38+00:00 2026-05-28T18:04:38+00:00

The ‘Wat’ talk for CodeMash 2012 basically points out a few bizarre quirks with

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The ‘Wat’ talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.

I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.

The behaviours specific to JavaScript (as I don’t know Ruby) are listed below.

I found in the JSFiddle that some of my results didn’t correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.

Empty Array + Empty Array
[] + []
result:
<Empty String>

I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video’s result.

Empty Array + Object
[] + {}
result:
[Object]

This matches the video’s result. What’s going on here? Why is this an object. What does the + operator do?

Object + Empty Array
{} + []
result:
[Object]

This doesn’t match the video. The video suggests that the result is 0, whereas I get [Object].

Object + Object
{} + {}
result:
[Object][Object]

This doesn’t match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.

Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN

Doing wat + 1 results in wat1wat1wat1wat1…

I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.

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  1. Editorial Team
    Editorial Team
    2026-05-28T18:04:39+00:00Added an answer on May 28, 2026 at 6:04 pm

    Here’s a list of explanations for the results you’re seeing (and supposed to be seeing). The references I’m using are from the ECMA-262 standard.

    1. [] + []

      When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.

    2. [] + {}

      Similar to [] + [], both operands are converted to primitives first. For “Object objects” (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).

    3. {} + []

      The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you’re not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.

    4. {} + {}

      Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:

      If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.

    5. Array(16).join("wat" - 1)

      As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that

      If either operand is NaN, the result is NaN.

      So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):

      If m is NaN, return the String "NaN".

      Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you’re seeing.
      When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").

    As to why you’re seeing different results for the {} + [] case: When using it as a function argument, you’re forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it’s instead parsed as an empty object literal.

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