Check JS variable for both not null and true

Question

What is the best way to check if a javascript variable is both not null and true?

So for example, let’s say I have the below code:

var trueVar = true; 
var falseVar = false; 

function checkNotNullAndTrue(someVar) {
    if (someVar != null && someVar) {
        return 1; 
    } else {
        return 0;
    }
}

checkNotNullAndTrue(trueVar) should return 1.

checkNotNullAndTrue(falseVar) should return 0.

checkNotNullAndTrue(someUndefinedVariable) should also return 0.

Is this the best way to do this or is there a better way?

use +!! operator like this +!!varname (it is three operators actually,+, !, !) — Bryan Chen
– Bryan Chen, Commented Mar 16, 2016 at 21:57
@BryanChen That is not cryptic enough. I am sure you can do better. — Sverri M. Olsen
– Sverri M. Olsen, Commented Mar 16, 2016 at 21:58
@SverriM.Olsen you need the ===+!!([!![]]+[]) synthetic operator if you want to just get true and not merely truthy. (Added this to my answer.) — djechlin
– djechlin, Commented Mar 16, 2016 at 22:08

djechlin · Accepted Answer · 2016-03-16 22:37:40Z

4

Kind of a weird question since null is falsy.

return x === true; // not null and 'true';
return x; // return truthy value if x not null and truthy; falsy otherwise
return !!x; // return true if x not null and truthy, false otherwise
return +!!x; // return 1 if x not null and truthy, 0 otherwise

!!x is same as !(!x) and casts x to true or false and negates, then negates a second time. Either a hack or a pattern that is same as Boolean(x) depending on your worldview.

+<boolean> will cast convert the boolean to the number 1 or 0.

Anyway someone requested a cryptic answer and "true" uses a lot of unnecessary characters so here it is:

return +!!(x === ([!![]]+[])); // 1 if x not null and true; 0 otherwise

edited Mar 16, 2016 at 22:37

answered Mar 16, 2016 at 21:58

djechlin

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Comments

Nina Scholz · Accepted Answer · 2016-03-17 08:41:06Z

Simply with the strict equality operator (===):

The identity operator returns true if the operands are strictly equal [...] with no type conversion.

function checkNotNullAndTrue(v) {
    return v === true ? 1 : 0;
}

or

function checkNotNullAndTrue(v) {
    return +(v === true);
}

Why hacky stuff does not work, sometimes:

// djechlin's part
write(+!!1);                           // 1 obviously not true
write(+!![]);                          // 1 obviously not true

// quentin's part
function test (valid_variable_name) {
    if (valid_variable_name) {
        return 1;
    }
    return 0;
}

write(test(1));                        // 1 obviously not true
write(test([]));                       // 1 obviously not true

// my part
var v = true;
write(+(v === true));                  // 1 true (the only one, that would work!)
write(+(1 === true));                  // 0 false, works
write(+([] === true));                 // 0 false, works


function write(x) {
    document.write(x + '<br>');
}

Quentin · Accepted Answer · 2016-03-16 21:56:17Z

2

Since null (and undefined, which you mention in your examples) are not true values, testing for them is redundant.

if (valid_variable_name) {
    return 1;
}
return 0;

… is sufficient.

… or if (valid_variable_name === true) if you want to test for true rather than any true value.

answered Mar 16, 2016 at 21:56

Quentin

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1 Comment

Nina Scholz Over a year ago

but on 1, you get 1 (true), it should return 0 (false).

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