JavaScript check if value is only undefined, null or false

I think what you're looking for is !!val==false which can be turned to !val (even shorter):

You see:

function checkValue(value) {
    console.log(!!value);
}

checkValue(); // false
checkValue(null); // false
checkValue(undefined); // false
checkValue(false); // false
checkValue(""); // false

checkValue(true); // true
checkValue({}); // true
checkValue("any string"); // true

That works by flipping the value by using the ! operator.

If you flip null once for example like so :

console.log(!null) // that would output --> true

If you flip it twice like so :

console.log(!!null) // that would output --> false

Same with undefined or false.

Your code:

if(val==null || val===false){
  ;
}

would then become:

if(!val) {
  ;
}

That would work for all cases even when there's a string but it's length is zero. Now if you want it to also work for the number 0 (which would become false if it was double flipped) then your if would become:

if(!val && val !== 0) {
  // code runs only when val == null, undefined, false, or empty string ""
}

The best way to do it I think is:

if(val != true){
//do something
} 

This will be true if val is false, NaN, or undefined.

One way to do it is like that:

var acceptable = {"undefined": 1, "boolean": 1, "object": 1};

if(!val && acceptable[typeof val]){
  // ...
}

I think it minimizes the number of operations given your restrictions making the check fast.

Well, you can always "give up" :)

function b(val){
    return (val==null || val===false);
}