Calling a function inside another function in JavaScript

Question

I have defined a function named larger to find the larger number between two arguments (num1, num2). Now I want to use this function inside another function called "largest" which gets an array and return the largest number of that array, but I`ve got stuck. Can anybody help me with that? Here is my codes:

function larger(num1, num2){
 var largerNumber = 0;
 if (num1 > num2){
 largerNumber = num1;
 } else {
 largerNumber = num2;
 }
return largerNumber;
}
function largest(array){
 for (var i = 0; i < array.length ; i++){
 for (var j = 0; j < array.length ; j++){
 if (array[i] != array[j]){
 //I don`t know if I am doing it right
 }
 }
 }
}

• ahitt6345: Are you trying to sort an array? What is the code supposed to do?
• Jaromanda X: your largest function will loop n squared times (where n is array.length) ... to get the largest value in an array only requires a loop n long - so, no you're not doing it right ... also, lookup Math.max for an even simpler solution
• Arunas: The easiest way to 'know if you're doing it right' is to try a simpler example, and see whether it works.

Answer 1

Just use Math.max:

function largest(array) {
 return Math.max.apply(Math, array);
}
console.log(largest([5,-2,7,6]));

If you really want to use a custom binary larger function, consider [].reduce:

function larger(num1, num2) {
 return num1 > num2 ? num1 : num2;
}
function largest(array) {
 return array.reduce(larger, -Infinity);
}
console.log(largest([5,-2,7,6]));

Answer 2

The simpliest solution is tu use a max tmp value. This way, you only need to do one iteration over all your array.

function largest(array){
 var max = array[0];
 for (var i = 1; i < array.length ; i++) { // So we start at 1
 max = larger(max, array[i]);
 // Or use this : if(array[i] > max) max = array[i];
 }

AS SAID BY ORIOL

I didn't check the length of the array, the above solution work only if the array.length> 0.

Otherwise, you'll have to check it usingsomething like this instead of var max = array[0];

function largest(array){
 var max = -Infinity; 
 for (var i = 0; i < array.length ; i++) { Start at 0
 max = larger(max, array[i]);
} 

It really depends on the ergonomics of your IHM.

• Oriol: Note the code assumes array.length > 0. Usually I prefer initializing max = -Infinity and starting the loop at i=0.
• Aks: Wep, didn't check this one. But I assume that OP will take care of this. (Ahah, i'll try to update my answers)
• Amir Omidvari: Thanks so much for your answer.

Answer 3

Iterate through the array once, keeping only the largest:

function larger(num1, num2){
 var largerNumber = 0;
 if (num1 > num2){
 largerNumber = num1;
 } else {
 largerNumber = num2;
 }
return largerNumber;
}
function largest(array){
 let largestNumber = array[0];
 for (var i = 1; i < array.length ; i++){
 largestNumber = larger(largestNumber, array[i]);
 }
 return largestNumber;
}
var test = [1, 53, 352, 22, 351, 333, 123, 5, 25, 96];
console.log(largest(test));

Answer 4

If you are just trying to find the maximal value, you should go with Oriol's answer: Math.max

If you are looking for a way to call one function from another you can do it like this:

function first_function() {
 //code of first function
 //call to the other function
 second_function();
}
function second_function() {
 //code of second function
}

Answer 5

Set the first element of array to a variable, if next element in array is greater set variable to that element; continue process, return variable

function largest(array) {
 if (!array.length) return;
 for (var i = 0, curr = void 0; i < array.length ; i++) {
 if (curr === void 0) curr = array[i];
 else if (curr < array[i]) curr = array[i];
 }
 return curr
}
console.log(largest([2,20,2000,2,7]));