strictEquals.js

strictEquals.js

// Your scientists were so preoccupied
// with whether or not they could,
// they didn't stop to think if they should.

// Like a === b
function strictEquals(a, b) {
  if (Object.is(a, b)) {
    // Same value.
    // Is this NaN?
    if (Object.is(a, NaN)) { // We already know a and b are the same, so it's enough to check a.
      // Special case #1.
      return false;
    } else {
      // They are equal!
      return true;
    }
  } else {
    // Different value.
    // Are these 0 and -0?
    if (
      (Object.is(a, 0) && Object.is(b, -0)) ||
      (Object.is(a, -0) && Object.is(b, 0))
    ) {
      // Special case #2.
      return true;
    } else {
      // They are not equal!
      return false;
    }
  }
}

My version:

function strictEquals(a, b) {
    if (Number.isNaN(a) && Number.isNaN(b)) return !Object.is(a, b);
    if ((Object.is(a, -0) ^ Object.is(b, -0)) && (Object.is(a, 0) ^ Object.is(b, 0))) return !Object.is(a, b);
  return Object.is(a, b);
}

Hope this is the shortest

function strictEquals(a,b){
  if(Number.isNaN(a) && Number.isNaN(b)) return false
  if(!Object.is(a,b) && Object.is(a+"",b+"")) return true
  return Object.is(a,b)
}

function stringEquals(a, b) {
    if (isNaN(a)) return false
    if ([0, -0].includes(a) && Object.is(a, -b)) return true
    return Object.is(a, b)
}

First special case NaN === NaN => both a and b will be NaN and therefore, and if either value is NaN, should return false. Other case, return the result of Objects.is(a, b)

Second sepecial case 0 === -0 or -0 === 0 => if the a is 0 or -0 and Object.is(a -b), its one of the special situations, that needs to return true. Otherwise, return Object.is(a, b)

I found it interesting that you managed to shorten if (isNaN(a) && isNaN(b)) return false to if (isNaN(a)) return false. You don't need to check if either value is NaN, you have enough with checking if the first value is NaN, in which case it will definitely evaluate to false. And you don't need to care about the remaining case in which b is NaN while a isn't, because Object.is() will handle it perfectly.

 function strictEquals(a,b){
     if(typeof a === 'number' && typeof b === 'number'){
         return a == b   
     }
     return Object.is(a,b)
 }

I couldn't figure it out on my own so I looked at your version and tried to minimize (not simplify!) it to the best of my ability

const strictEquals = (a,b) => (Object.is(a,b) 
     ? !Object.is(a, NaN) 
     : (Object.is(a, 0) && Object.is(b, -0)) || (Object.is(a, -0) && Object.is(b, 0)))
}

function strictEquals(a, b) {
    if(Object.is(a, NaN) && Object.is(b, NaN)) return false;
    if(Object.is(a, -0) && Object.is(b, 0)) return true;
    if(Object.is(b, -0) && Object.is(a, 0)) return true;
    return Object.is(a, b);
}

function strictEquals(a,b){
    if(Number.isNaN(a)&&Number.isNaN(b)){
        return false;
    }
    if(Number.isFinite(a) && Number.isFinite(b) && Math.abs(a)===0 && Math.abs(b)===0){
        return true;
    }
    return Obejct.is(a,b);
}

const strictEquals = (a, b) => {
        if (Object.is(a, NaN) && Object.is(NaN, b)) {
                return false; 
        } else if (Object.is(a, 0) && Object.is(a, -0)) {
                return true 
       } else if (Object.is(a, -0) && Object.is(b, 0)) {
                return true 
        } 
            return Object.is(a,b) 
        
}

const strictEquals = (a, b) => {
  const isSameValue = Object.is(a, b)
  // === behaves as same as Object.is() checks the both args are of same value or not
  // edge cases 
  // 0 === -0 || -0 === 0 // true
  // NaN === NaN // false 
  
  if(isSameValue && Object.is(a, NaN)){
      return false
   }
  if((Object.is(a , 0) && Object.is(b, -0)) || (Object.is(a, -0) && Object.is(b, 0))){
    return true
  }
  
 return isSameValue
}