Discussion on: The Power of Higher Order Functions in JavaScript (With examples and use cases)

[SerjoA]

i also suggest a re write to this functionality of the last pass

filterMaleFrogs = function (frog) {
   return frog.gender === 'Male'
};

filterAdultFrogs = function (frog) {
   return frog.age >= 10
};

filterFrogNamesThatStartWithHippo = function (frog) {
   return frog.name.toLowerCase().startsWith('hippo')
};

filterGmailEmails = function (frog) {
   return /gmail.com/i.test(frog.email)
};

function composeFrogFilterersModified(...fns) {

   return function (frogs) {
      return frogs.filter(
         combineFilters(fns)
      );
   }
}

function combineFilters(fns) {
   return function (frog) {
      return fns.reduce(function (accumulatedFilter, currentFilter) {
         if(accumulatedFilter === false) {
            return false;
         }

         return !!currentFilter(frog);

      }, true);
   }
}

const applyFrogFilterersComposedAgain = composeFrogFilterersModified(
   filterMaleFrogs,
   filterAdultFrogs,
   filterFrogNamesThatStartWithHippo,
   filterGmailEmails,
);

const allFilteredFrogsModified = applyFrogFilterersComposedAgain(combinedFrogsList);
console.log('all filtered frogs: ', allFilteredFrogsModified);

from my understanding, this will only go once per frog and not create array.filter 4 times
please correct me if iam wrong or missed something on your implementation

[jsmanifest]

Hi SerjoA, thank you for the suggestion and the workaround!

I will edit the post and swap out the current code example with the example you provided because its correct.

I'm going to make a slight modification to your combineFilters function to run this instead:

function combineFilters(...fns) {
   return function (frog) {
     for (let i = 0; i < fns.length; i++) {
       const filter = fns[i]
       const passes = filter(frog)
       if (passes) {
         continue
       } else {
         return false
       }
     }
     return true
   }
}

This is a slightly more optimized version which reduces the amount of iterations as much as possible.

Thank you!

[SerjoA]

great, looks good