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\'

Answer 1

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 2

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 3

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 4

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

Like multiply([1,2,3])

Demo here

Answer 5

You're not calling multiply as a function:

multiply([1,2,3]);

Answer 6

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