Neatest way to check the parity of a BigNumber

[PaulRBerg]

I did this:

if (x.mod(2).eq(BigNumber.from(0))) {
  // Even number
} else {
  // Odd number
}

But it feels somewhat verbose. Is there a shorter/ neater way?

[ricmoo]

Keep in mind == doesn't work on BigNumbers, you would need x.mod(2).eq(BigNumber.from(0)).

But either of these is prolly better:

• x.mod(2).isZero()
• x.mod(2).eq(0)
• x.and(1).isZero()
• x.and(1).eq(0)

Hope that helps. :)

[PaulRBerg]

Cheers!