Discussion on: We don't need a ternary operator

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Replies for: LOL I feel the exact same way. The Ternary feels like an in-between. Somewhere in the middle of imperative and functional. Today I was actually wo...

The problem with this is that both orderCountText and noOrderCountText will be evaluated in either case.
Do they have to be functions?

JavaScript Joel • Sep 7 '18

The problem with this is that both orderCountText and noOrderCountText will be evaluated in either case.

only one function will execute in this case, either the if case or the else case, but never both.

In my version of ifElse, they can be values that will be returned or functions to be executed.

Mihail Malo • Sep 8 '18

Yeah, whereas in condition ? expensiveGet() + 1 : expensiveGet2() - 1 only one expression will be evaluated.

JavaScript Joel • Sep 8 '18

Same with ifElse

JavaScript Joel • Sep 8 '18 • Edited

const expensiveGetPlus = x => y => expensiveGet(y) + x

ifElse (condition) (expensiveGetPlus (1)) (expensiveGetPlus (-1)) (true)

JavaScript Joel • Sep 8 '18

Check out Sanctuary's ifElse sanctuary.js.org. You can live edit their page to see how it works.

I'd love to get some live editing docs page for my stuff too. Pretty neat site.

Mihail Malo • Sep 8 '18

I see it's actually for threading through the value, just like tap:

S.ifElse(x=>typeof x === "number")(x=>x-1)(x=>x+"a")(3)
 2

S.ifElse(x=>typeof x === "number")(x=>x-1)(x=>x+"a")("banana")
 "bananaa"

However, for normal conditional use it's error prone, as you've shown in your code. Both expensive gets will run before the condition is even applied to true.
The actual code would be

const expensiveGetPlus = x => () => expensiveGet() + x
const expensiveGet2Plus = x => () => expensiveGet2() + x

ifElse (condition) (expensiveGetPlus (1)) (expensiveGet2Plus (-1)) (true)

JavaScript Joel • Sep 8 '18

That's correct, ifElse accepts functions. So those functions are only executed when the condition is met.

JavaScript Joel • Sep 8 '18

Both expensive gets will run before the condition is even applied to true.

This is still false. Only one will run. Never both.

const isEven = n => n % === 0
const logEven = n => console.log(`${n} is Even!`)
const logOdd = n => console.log(`${n} is Odd`)

ifElse (isEven) (logEven) (logOdd) (10)

//=> "10 is Even!"

You can see logOdd is never called.

Mihail Malo • Sep 9 '18

In your first example

const expensiveGetPlus = x => expensiveGet() + x

ifElse (condition) (expensiveGetPlus (1)) (expensiveGetPlus (-1)) (true)

both run.

const expensiveGet = () => {
  console.log("doing expensive get")
  return 2
}

JavaScript Joel • Sep 9 '18

Yes you are correct. The expensiveGet method needs to take 2 arguments for it to work the way it is being called in ifElse.

I have created a working example that you can run.

const S = require('sanctuary')

const expensiveGet = (y) => {
  console.log("doing expensive get for", y)
  return 2
}
const condition = x => x === 100
const expensiveGetPlus = x => y =>
  expensiveGet(y) + x

S.ifElse (condition) (expensiveGetPlus (1))
  (expensiveGetPlus (-1)) (100)
//=> "doing expensive get for 100"

This example will show that ifElse only executes one of the functions.