Products of all odd integers

JavaScript
1

Hello ,
back at it again…
After multiple tries and a lot of frustration I managed to figure out this code.

function oddProduct(arr) {
	return arr.filter( x => (x % 2 == true) ).reduce( (a,b) => a*b ,1 );
	}

This is supposed to multiply all the odd integers in an array and return that value.
So my question is…

When I used arr.filter, I tried using (x => (x % 2 == 0)) first, but realized that would give me EVEN numbers, so then I tried “true” and “voila” it worked… So can someone help me understand why having (x % 2 == true) ) works and returns the odd numbers?

Is it because true integers are 1,2,3,4, etc… which means there IS a remainder meaning it is an odd number?

Also when I use the reduce function , reduce( (a,b) => a*b ,1 ) how would the variables (a,b) know which numbers to multiply?

Thanks for all the help,
Peter

2

it’s because 1 == true is true

you have x % 2 that would give or 0 for even numbers and 1 for odd numbers - you could use x % 2 === 1 and it would work and be more readable

but you have used x % 2 == true which works with the loose equality operator because 1 and true can be converted in each other with type coercion

tldr: it’s becauze javascript is weird

3

I would suggest you use === instead of ==. In this instance it would force you to think more about what the % operator returns and what you are actually looking for from that return value. With == you are depending on the JS interpreter to do some conversions behind the scenes, which I guess is fine if you really understand what that implies, but often it can cause bugs on edge cases.

As @ILM said, JS is weird. Using strict equality will take all of that weirdness out of the equation.

4

Yeah if I were to use x % 2 === 1that would make more sense. Thank you.
On another note,

When I use the reduce function , reduce( (a,b) => a*b ,1 ) how would the variables (a,b) know which numbers to multiply?

Thanks,
Peter

5

Yeah, as @ILM stated that does seem to help me understand it a bit more. On another note, do you happen to know how the reduce function works?

For example when I use the reduce function , reduce( (a,b) => a*b ,1 ) how would the variables (a,b) know which numbers to multiply ? Does it work like the filter function and assign new variables to the array element? And does it execute the function until there are no more elements?

Much appreciated,
Peter

6

the variables don’t know anything

the reduce method call the function you pass in as argument with different values each time. The first parameter a is the accumulator, the second is the current value

yoo could write it as (product, number => product * number to havemore descriptive names for tbe function parameters

see the documentation:

7

Ah okay , I was confused since I thought the accumulator added the elements together but now I realize you can call accumulator * currentValue … Okay thanks, will look for some videos so that I can understand it fully.

-Peter