Why check for !isNaN() after isFinite()?

Question

The only difference is this:

!isNan(1/0) // --> true
isFinite(1/0) // --> false

isNaN checks whether the argument is a number or not. The Infinities (+/-) are also numerical, thus they pass the isNaN check, but don't pass the isFinite check.

** Note that any string which can be parsed as a number ("2", "3.14") will cause isNaN to return false.

Hope this helps.

PS: The answer given by user1170379 was very nearly perfect.

you might reason out [Why?] after reading this:

NaN doesn't check if the passed value is infinite or not - it checks if the input val evaluates into a "Type: Number" end-result. Because isNaN(string) is accepted, so the: isNaN("3.14") //false (which means true, the given token is duck converted into a type Number successfully )

You may understand that the input value may happen to be an unresolved brute number, even a math operation as simple as: (x/y); which in turn might yield a (+/-infinity) number.

Here x=1, y=0; meaning (1/0).Then isNaN(x/y) will first evaluate to isNaN(1/0); then to isNaN(infinity) //false. Since (1/0)=infinity is of type: "number" ie typeof(1/0) //"number" isNaN should and will return false.

You don't want to put "infinity" where an end result number is expected.

Probably for the same reason that I have implemented (isfinite(num) && isfinite(-num)) - I was getting errors from mysql complaining about putting "-nan" into the database even though I had a check for isfinite(field)...

A useful article on this subject is http://jacksondunstan.com/articles/983 which provides an optimization ((d*0.0)==0.0)

isNaN() returns true if the argument is not a number or if the argument is a non-numeric value such as a string or an object.Otherwise, It returns false. Example: isNaN(0/0) =>true; isNaN(2-1) =>false; isFinite() returns true if the argument is a number other than NaN,Infinity or -Infinity.Otherwise, It returns false. Example: isFinite("2000") =>false; isFinite(200/2) =>true;`