Check exactly one boolean option set

Question

Well, this is kind of hacky:

function b2n(boo) {
    return boo ? 1 : 0;
}

if(b2n(opt1) + b2n(opt2) + b2n(opt3) !== 1) {
    throw new Error(\"Exactly one o         


        

Answer 1

You mentioned in your comment that this is coming from a commander options object.

You can do this more elegantly using Lodash:

if (_(options).values().compact().size() === 1)

If you only want to count a subset of the options, you can insert

.pick('a', 'b', 'c')

Answer 2

if ([opt1, opt2, opt3].reduce(function(x, y) { return x + !!y }, 0) == 1) {
    // exactly one
};

ECMAScript 5 reduce function.

Answer 3

Assuming you had an array of options, you could do:

if(opts.filter(Boolean).length !== 1) {}

It seems to me though that you ought to have one variable with three possible states instead...

var opt = 'a'; // (or 'b', or 'c')

Answer 4

I think you are being too clever, what's wrong with:

var optionsSelected = 0;
if( opt1 ) optionsSelected++;
if( opt2 ) optionsSelected++;
if( opt3 ) optionsSelected++;

if( optionsSelected !== 1 ) {
    throw new Error("Exactly one option must be set");
}

Of course I can play the clever game too:

 if( opts.filter(Boolean).length !== 1 ) {
     throw new Error("Exactly one option must be set");
 }

Answer 5

You can do this :

if ( !!opt1 + !!opt2 + !!opt3 !== 1 ) {

It works because

• !! makes a boolean from any value (true if the objects evaluates as true in if(value))
• when adding booleans you get 1 for true and 0 for false.

Answer 6

@spudly is on the right track, but it could be a little more compact:

if( [opt1,opt2,opt3].filter(function(x){return x}).length!==1 ) {
    throw new Error("Exactly one option must be set");
}

See ES5's filter method for more information.