javascript - Why does Math.min([]) evaluate to 0?

This happens because [] is
coerced to 0.

You can see this
with the following call:

(new
            Number([])).valueOf(); //
            0

Therefore, calling
Math.min([]) is the same as calling
Math.min(0) which gives
0.

---

I
believe that the reason that new Number([]) treats
[] as 0 is
because:

1. The href="http://www.ecma-international.org/ecma-262/6.0/#sec-number-constructor-number-value"
   rel="nofollow noreferrer">spec for the Number(value)
   constructor uses a ToNumber
   function.


2. The href="http://www.ecma-international.org/ecma-262/6.0/#sec-tonumber" rel="nofollow
   noreferrer">spec for the ToNumber(value) function
   says to use ToPrimitive for an object
   type (which an array is).


3. The primitive value of an array
   is equal to having the array joined, e.g. [] becomes
   "", [0] becomes
   "0" and [0, 1] becomes
   "0,1".


4. The number constructor
   therefore converts [] into "" which is
   then parsed as
   0.

The
above behaviour is the reason that an array with one or two numbers inside it can be
passed into Math.min(...), but an array of more
cannot:

• Math.min([])
  is equal to Math.min("") or
  Math.min(0)


• Math.min([1])
  is equal to Math.min("1") or
  Math.min(1)


• Math.min([1,
2]) is equal to Math.min("1,2") which cannot be
  converted to a number.