Closed cherche closed 8 years ago
is there no if.morethanonecharacter or something?
Well, it would be if(x.length > 1), but that's not very useful in checking if the number is an integer.
if(x.length > 1
This was a mistake on my end. I was supposed to use !isNaN(y) instead of y !== NaN.
!isNaN(y)
y !== NaN
is there no if.morethanonecharacter or something?