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

2018-06-16 23:44:24

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Why does ++[[]][+[]]+[+[]] return the string “10”? 8 answers

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

Because the spec says so:

Math.min casts each parameter to a number using...

ToNumber casts objects to a number using...

ToPrimitive casts objects to primitive values using...

[[Default Value]] internal method converts objects to primitives based on their hint parameter.

The default hint for all objects is string. Which means the array gets converted to a string, which for [] is "" .

ToNumber then converts "" to 0 , per the documented algorithm

Math.min then takes the only parameter and returns it, per its algorithm.

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:

The spec for the Number(value) constructor uses a ToNumber function.

The spec for the ToNumber(value) function says to use ToPrimitive for an object type (which an array is).

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

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.

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