CODEWITHSUNDEEP
javascriptarraysloopsfor-loopmultidimensional-array
Zr Classic
Zr Classic

Reputation: 313

can't return value/variable in a function

when I console.log it display all the arrays to be the string, because in my arrays, they are multi-dimensions; look below the code using console.log in function, without return but undefined

function changeStrings(arr, replacement) {
    var newArr = []

  for ( var i = 0 ; i < arr.length ; i++) {
    newArr.push(arr[i].split(" "))
  }

  for ( var j = 0 ; j < replacement.length ; j++ ) {
    newArr[0][3] = replacement[j-2]
    newArr[1][3] = replacement[j-1]
    newArr[2][4] = replacement[j]
  }
   
  for ( var k = 0 ; k < newArr.length ; k++ ) {
    console.log(newArr[k].join(" "))
  }
 
};


let initial = ["my city in London", "my name is Mike", "my phone number is 00909090"];
let replacements = ['Paris', 'John', '1234'];

console.log(changeStrings(initial, replacements))

and if i using return , it just print one line;

check this 

function changeStrings(arr, replacement) {
    var newArr = []

  for ( var i = 0 ; i < arr.length ; i++) {
    newArr.push(arr[i].split(" "))
  }

  for ( var j = 0 ; j < replacement.length ; j++ ) {
    newArr[0][3] = replacement[j-2]
    newArr[1][3] = replacement[j-1]
    newArr[2][4] = replacement[j]
  }
   
  for ( var k = 0 ; k < newArr.length ; k++ ) {
    var dispplay = newArr[k].join(" ")
    return dispplay
  }
 
};

// now let's test out our functions!
let initial = ["my city in London", "my name is Mike", "my phone number is 00909090"];
let replacements = ['Paris', 'John', '1234'];

console.log(changeStrings(initial, replacements))

the newArr variable is array multidimension it looks like this ;

[ [ 'my', 'city', 'in', 'Paris' ],
  [ 'my', 'name', 'is', 'John' ],
  [ 'my', 'phone', 'number', 'is', '1234' ] ]

i am trying to using for loop for this case, because i want to train my logic;), can anyone help me to find out what's wrong ? or fix this ? thank you

Upvotes: 1

Views: 54

Answers (2)

HynekS
HynekS

Reputation: 3307

The problem is that return statement end the whole loop immediately (same as break) . You can not return directly like that, if you want to finish the loop. On the other way, you can use it purposefuly to end the loop, for instance if you found a desired item and don't neet to loop anymore.

How to actually use return in loop:

let fruit = ['apple', 'orange', 'pear', 'strawbery'];
let myFavoriteFruit === 'orange';

const IncludesMyFavouriteFruit = (list) => {
    for (let item of list) {
        if (item === myFavoriteFruit) return true;  // Ends the loop; we'll never get to 'pear' and 'strawbery'.
    }
    return false; // There is not my favourite fruit.
}

Sorry for using ES6, too lazy to write old JS.

Upvotes: 0

Mamun
Mamun

Reputation: 68933

The function is returning just after the first iteration of the loop. That prevents the rest of the iterations from execution. You have to return an array from the function. You can push() all the item to the array inside the loop then return the array:

function changeStrings(arr, replacement) {
  var newArr = []

  for ( var i = 0 ; i < arr.length ; i++) {
    newArr.push(arr[i].split(" "))
  }

  for ( var j = 0 ; j < replacement.length ; j++ ) {
    newArr[0][3] = replacement[j-2]
    newArr[1][3] = replacement[j-1]
    newArr[2][4] = replacement[j]
  }
  var display = [];
  for ( var k = 0 ; k < newArr.length ; k++ ) {
    display.push(newArr[k].join(" "));
  }
 
  return display
};

// now let's test out our functions!
let initial = ["my city in London", "my name is Mike", "my phone number is 00909090"];
let replacements = ['Paris', 'John', '1234'];

console.log(changeStrings(initial, replacements))

Upvotes: 1

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