Javascript compare 3 values

Works for any number of items.

ES5

if ([f, g, h].every(function (v, i, a) {
  return (
    v === a[0] &&
    v !== null
  );
})) {
  // Do something
}

ES2015

if ([f, g, h].every((v, i, a) => 
  v === a[0] &&
  v !== null
)) {
  // Do something
}

What about

(new Set([a,b,c])).size === 1

You could shorten that to

if(g === h && g === f && g !== null)
{
//do something
}

---

For an actual way to compare multiple values (regardless of their number)
(inspired by/ simplified @Rohan Prabhu answer)

function areEqual(){
   var len = arguments.length;
   for (var i = 1; i< len; i++){
      if (arguments[i] === null || arguments[i] !== arguments[i-1])
         return false;
   }
   return true;
}

and call this with

if( areEqual(a,b,c,d,e,f,g,h) )
{
//do something
}