Multiply all elements in array

Question

I couldn't find an example here what I'm really looking for. I'd like to multiply all array elements, so if an array contains [1,2,3] the sum would be 1 2 3=6; So far I've got this code, but it returns undefined.

function multiply (array) {
    var sum=1;
    for (var i=0; i<array.length; i++) {
        sum = sum * array[i];
    } 
    return sum;
}
console.log(multiply[1,2,3]);

Could anyone please explain, what am I missing here?

Answer 1

The cause is already known. Here's an alternative - using Array.reduce for your method:

 console.log( [1, 2, 3].reduce( (a, b) => a * b ) ); console.log( Array.from( {length: 20} ) .map( (v, i) => i + 1 ) .reduce( (a,b) => a * b ) .toLocaleString()); // for empty arrays, use some initial value const arr = []; if (arr.reduce( (a, b) => a * b, -1 ) === -1) { console.error(`The given array ${arr} is empty`); }

See also

Answer 2

The alternative way to use Array.reduce should have the initial value set to 1 or else our function will return 0 no matter what.

[1, 2, 3].reduce((a, b)=> a*b, 1)

per https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

Answer 3

You need to have () when you call the function.

Like multiply([1,2,3])

Demo here

Answer 4

您没有将 multiply 作为函数调用：

multiply([1,2,3]);

Answer 5

This is the simplest way to do any arithmetic operation on array

 myMathArray = [1,2,3,4,5] let myValue = myMathArray.reduce((a,b)=> a*b) console.log(myValue)

Answer 6

If you want to multiply some consecutive numbers like 1,2,3 then, then apply this code and enter the number in the console (arr)

let array = [];

function factorisation(arr) {
    for (let j = arr; j > 0; j--) {
        array.push(j);
    }
    return multiply();
}

function multiply() {
    var sum = 1;
    for (var i = 0; i < array.length; i++) {
        sum = sum * array[i];
    }
    return sum;
}

console.log(factorisation(5));
//5*4*3*2*1 = 120

Answer 7

Using Lodash >=4.7:

_.reduce(array, _.multiply)

or, using all three arguments:

_.reduce(array, _.multiply, 1)

Answer 8

The reduce() method executes a provided function for each value of the array and reduces the array to a single value

 const arr = [1,2,3]; const sum = arr.reduce((prevValue,curValue) => { return prevValue * curValue },1); console.log(sum);

prevValue is the initial value and it's equal 1 in this example

curValue is the value of the current element in array

Answer 9

This can be simplified using recursion...

    function productOfArray(arr) {

    let sum = 1;
    
    function sumHelper(nums) {

        if(nums.length === 0)
        return 0;

        sum = sum * nums[nums.length-1];
        nums.length--;
        
        sumHelper(nums);
        
    }

    sumHelper(arr);
    
    return sum;
    
}

console.log(productOfArray([1,2,3])); 

Answer 10

If you want to use recursion

 const productOfArray = (arr) => { if (arr.length === 0) { return 1; } return arr[0] * productOfArray(arr.slice(1)); } console.log(productOfArray([1,2,3,10])); // 60

Answer 11

multiply all element array use below example:

const arr=[1,2,3,2]

use reduce();

const multiply=arr.reduce((preValue,nextValue)=>preValue *nextValue )

console.log(multiply)//12