I am writing a function that will return the first n elements of an array and if n is undefined then return the first element.
*Edit: I solved the problem.
_.first = function(array, n) {
var result = [];
if (n == undefined) {
return array[0];
}
var m;
if (array.length < n) {
m = array.length;
} else {
m = n;
}
for (var i = 0; i < m; i++) {
result.push(array[i]);
} return result;
};
This program is basically checking if the n value is bigger than the array's length, if it is, then it exits.
If n is not a number it exits. If it is it executes the program and logs the values of indexes with for loop until i reaches the n value. Also, it pushes the values to the empty array, so you can get the values from the array for later use.
var arr1 = [2, 3, 4, 5, 6, 7, 8];
var arr2 = []; //empty array
function arrNreturn(arr, n){
if(typeof n != 'number'){
console.log('n is not a number');
return false; //exit the program
}
if(n > arr.length){
console.log("the n value is bigger than the length");
}else{
for(var i = 0; i < n; i++){
console.log(arr[n]);
arr2.push(arr[n]);
}
}
}
arrNreturn(arr1, 10);
1_.first = function(array, n) {
2 var result = [];
3 for (var i = 0; i < n; i++) {
4 result.push(array[i]);
5 return result;
6 } return array[0];
7};
the problem is on the 5th line, you seem to be returning the first and not taking into account the n parts
a solution might be
_first = function(array, n){
var result = [];
if(n === undefined) return array[0];
for (var i = 0; i < n; i++) {
result.push(array[i]);
}
return result;
}
Basic is
const newArr = function(arr) {return n }
Shorthand
const newArr = arr => { return n}
array.slice(0, n)?