finding the whole number portion of a number

Question

Using Math.floor:

Math.floor(number)

You can also use newNumber = parseInt(number, 10);

Bitwise operators - the shortest syntax for positive numbers having integral part of number <= 2^31-1

~~1.2 // 1
~~1.5 // 1
~~1.9 // 1

1.2>>0 // 1
1.5>>0 // 1
1.9>>0 // 1

1.2|0 // 1
1.5|0 // 1
1.9|0 // 1

With values exceeding 2^31-1 will return incorrect results.

With ES6, there is Math.trunc(), this function returns the integer part of a number by removing any fractional digits.

This differs from the Math.floor() while handling the negative numbers.

console.log(Math.trunc(3.14)); // 3

console.log(Math.trunc(-3.14)); // -3
console.log(Math.floor(-3.14)); // Here floor gives -4.

So essentially you can get rid of writing this longer expression

number > 0 ? Math.floor(number) : Math.ceil(number)

for negative numbers you can just use Math.abs(num) and it will knock off the - sign from the start