Trick to call a function inside tenary expression and use its return-value without calling it again

Question

I'm looking for a neat way to do this:

return func(arr) >= 0 ? func(arr) : arr.length;

Calling func(arr) does not modify anything, so the above works fine.

But I would still like to avoid executing func(arr) twice.

Is there a neat way to achieve that in JavaScript?

@deceze: That would return `true` or `false`, most certainly not `arr.length` (let alone the fact that when `func(arr)` returns -1, it will return -1 instead of `arr.length`). — goodvibration, Mar 04 '20 at 10:46
Well, you are wrong in assuming that `||` returns a boolean. True about the `-1` though, you'd need to clarify @Nina's request there. — deceze, Mar 04 '20 at 10:47
@deceze: Clarify that -1 < 0? I assumed that the fact that I was using `>= 0` and not `!= 0` would make that pretty obvious. — goodvibration, Mar 04 '20 at 10:49
Fine, then something like `(i => i >= 0 ? i : arr.length)(func(arr))` is about as good as it gets. See the duplicate for further options. — deceze, Mar 04 '20 at 10:51
@deceze: Yes, thought about it, but wasn't sure it would work. Anyway, this suggestion appears in one of the answers to one of the two questions suggested as duplicate of mine. Thanks!!! — goodvibration, Mar 04 '20 at 10:54
Frankly though, before writing that sort of abomination, I'd do a simple `const foo = func(arr); if (foo >= 0) return foo; return arr.length;`… — deceze, Mar 04 '20 at 10:58
If `func` always returns `-1` or an integer greater than or equal to zero, you can use this: `return ~(~func(arr)||~arr.length);`. Demo: https://jsfiddle.net/5vf3L78m/ — Unmitigated, Mar 18 '20 at 20:48

score 0 · Answer 1 · answered Mar 04 '20 at 10:45

0

Try:

let retVal = func(arr);
return retVal  >= 0 ? retVal : arr.length;

answered Mar 04 '20 at 10:45

JohnPan

1 Answers1