Prevent a boolean variable returning to false in JavaScript

Question

I just want to check I'm doing this right. I have a variable bar that starts false and is set true if a function foo() returns false, however I check foo() against several arguments and don't want bar returning to false if it is ever set true. This is what I've written:

var bar = false;
var collection = []; // this is filled with arguments for foo
for(var i = 0; i < collection.length; i++) {
    bar = !foo(collection[i]) || bar;
}

Should that do the trick? Or is there maybe a simpler way?

Does "foo()" have side-effects that you want to happen for each member of the collection? Also, don't forget var for the variable "i" in the loop! — Pointy
– Pointy, Commented Mar 25, 2011 at 13:40
The question is a bit confusing, but by having bar in an OR like that you do guarantee that once true it will stay true. Both of pointy's points are good ones - mind, bar will most likely be out of foo's scope. — idbentley
– idbentley, Commented Mar 25, 2011 at 13:40
When you say "returning to false," do you mean "being set to false"? Variables do not return, only functions do. — Matt Ball
– Matt Ball, Commented Mar 25, 2011 at 13:40
Yes, I want foo() to execute for each array element. Have added var i! Also, yes I meant not be set to false after being set to true. — tamewhale
– tamewhale, Commented Mar 25, 2011 at 13:44

Richard Dalton · Accepted Answer · 2011-03-25 13:40:08Z

3

Change your for loop to do this and it should jump out as soon as it is set to true:

var bar = false;
var collection = []; // this is filled with arguments for foo
for(i = 0; i < collection.length && !bar; i++) {
    bar = !foo(collection[i]) || bar;
}

answered Mar 25, 2011 at 13:40

Richard Dalton

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4 Comments

Pointy Over a year ago

Don't forget the var declaration for "i"!

tamewhale Over a year ago

Yeah that's a good idea. Unfortunately in this case I need foo() to execute for every array element regardless.

Richard Dalton Over a year ago

@tamewhale In that case your code works fine. You can test things like this easily on jsfiddle - jsfiddle.net/s6qN3

pimvdb Over a year ago

I would swap the || part - the function is always executed. Unless you find this necessaty, bar || !foo... can be faster, depending on what kind of expensive things foo does.

Philippe · Accepted Answer · 2019-01-01 21:33:00Z

0

I often use Array.maps() and Array.includes() for such things.

Array.map() runs a function on each item in the array and returns the value into a new array.

Array.includes() tests an array to see if it contains a value.

let collection = []; // this is filled with arguments for foo
let test = collection.map((currentItem) => {
  return foo(currentItem);
});
return test.includes(true);

answered Jan 1, 2019 at 21:33

Philippe

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