正确使用array.map

Question

I am not sure why my array.map function returns undefined but my for loop returns the result. When I console log map it displays:

Tom Dick Harry

and when console log for loop it shows

['Tom', 'Dick', 'Harry']

Not sure if this is the problem of showing undefiend.

var friends = ["Tom", "Dick", "Harry"];
var secondLevelFriends = ["Anne", "Harry", "Quinton"];
var allUsers = ["Tom", "Dick", "Harry", "Anne", "Quinton", "Katie", "Mary"];

function findPotentialFriends(existingFriends) {
  return function(x) {
    // existingFriends.map(function(elem) {
    //   if(elem === x) return false;
    //   else return true;
    // })
    for(var i = 0; i < existingFriends.length; i++) {
      if(existingFriends[i] === x) return false;
      else return true;
    }
  }
}

var isNotAFriend = findPotentialFriends( friends );
isNotAFriend(allUsers[0]); // false

Answer 1

First of all your for loop has some logic problems. it should be written as so:

var foundFriend = false;
for(var i = 0; i < existingFriends.length; i++) {
  if (existingFriends[i] === x)
  {
    foundFriend = true;
    break;
  }
}
return !foundFriend;

The reason why isNotAFriend evaluates to undefined is because you forgot a return statement. Furthermore, the Arrays.map just simply takes each element of the array, passes it into the function, then creates a new array based on what your function returns. What you might be searching for is perhaps Array.some see docs . Example:

 var friends = ["Tom", "Dick", "Harry"]; var secondLevelFriends = ["Anne", "Harry", "Quinton"]; var allUsers = ["Tom", "Dick", "Harry", "Anne", "Quinton", "Katie", "Mary"]; function findPotentialFriends(existingFriends) { return function(x) { return !existingFriends.some(function(elem) { return elem === x; }); } } var isNotAFriend = findPotentialFriends( friends ); console.log(isNotAFriend(allUsers[0])); 

Answer 2

The problem with your code is that the return statements in the callback just return to map() , they don't return to the original caller. map() collects all these return values into an array and returns this, but you never do anything with the result of existingfriends.map() .

Javascript has a built-in function to search an array for a matching element, Array.prototype.indexOf() . It returns -1 if no match is found.

So you can simply write:

 var friends = ["Tom", "Dick", "Harry"]; var secondLevelFriends = ["Anne", "Harry", "Quinton"]; var allUsers = ["Tom", "Dick", "Harry", "Anne", "Quinton", "Katie", "Mary"]; function findPotentialFriends(existingFriends) { return function(x) { return existingFriends.indexOf(x) == -1; } } var isNotAFriend = findPotentialFriends(friends); console.log(isNotAFriend(allUsers[0])); // false 

Answer 3

Simply add the return in map() function .Its one of the new function inside the return function .so apply with return this also.

 var friends = ["Tom", "Dick", "Harry"]; var secondLevelFriends = ["Anne", "Harry", "Quinton"]; var allUsers = ["Tom", "Dick", "Harry", "Anne", "Quinton", "Katie", "Mary"]; function findPotentialFriends(existingFriends) { return function(x) { return existingFriends.map(function(elem) { if(elem === x) return false; else return true; }) // for(var i = 0; i < existingFriends.length; i++) { // if(existingFriends[i] === x) return false; // else return true; //} } } var isNotAFriend = findPotentialFriends(friends); console.log(isNotAFriend(allUsers[0])); // false