Array.reduce(): Addition
A single value from an array can be calculated using the reduce() function. In other words, an array is condensed into a single value.
Example:
const numbers = [20, 13, 5, 31];
const sum = numbers.reduce((accumulator, current) => {
return accumulator + current;
}, 0);
Let’s examine it step by step:

We call the
.reduce()
method on thenumbers
array. 
We assign the result of
numbers.reduce()
to a new variable calledsum
. 
The reduce method needs two parameters: the reducer and the starting value.
Reducer
Example:
(accumulator, current) => {
return accumulator, current;
}
This is a callback that is applied to each element in the array, but this callback has two parameters: accumulator
and current
.
The last value that the reduce function last calculated is always referenced by the accumulator
. Additionally, the term current
refers to a single element in the array.
Example:
const numbers = [20, 13, 3, 31];
const sum = numbers.reduce((accumulator, current) => {
return accumulator + current;
}, 0);
Detailed Explanation:
The first time the callback is executed, accumulator is set to 0 (due to the initial value) and current is set to the first element of the array. Consequently, accumulator is 0 and current is 20.
Then we output accumulator + current
, which is 0 + 20 = 20. The updated value of the accumulator
is 20.
The callback executes a second time, with current = 13
(the second item in the array) and accumulator = 20
. We return accumulator + current
, which is 20 + 13 = 33
. The accumulator
is 33.
The callback executes a third time, with current = 3
(the third item in the array) and accumulator = 33
. We return accumulator + current
, which is 33 + 3 = 36
. The accumulator
is 36.
The callback executes a fourth time, with current = 31
(the fourth item in the array) and accumulator = 36
. We return accumulator + current
, which is 36 + 31 = 67
. The accumulator
is 67.
The array has no additional items, therefore the reduce produces the final value of accumulator
, which is 67.
Inital Value
Reducer
and initialValue
are the two parameters that the .reduce()
method accepts (not to be confused with the 2 parameters of reducer
i.e. accumulator
and current
). The value we assign to the accumulator
the first time the callback executes is known as the initialValue
.
The first time that callback is executed, JavaScript will automatically take your initialValue
and pass it to the accumulator
parameter in the reducer
.
Array.reduce(): Multiplication
Example:
const numbers = [1, 2, 5];
const finalResult = numbers.reduce((accumulator, current) => {
return accumulator * current;
}, 1);
console.log(finalResult); // 10
The initial value for multiplication
We cannot begin a multiplication with a value of 0
. This is due to the fact that any number multiplied by 0 will produce 0, 1 * 0 = 0
. Given that any number multiplied by 1 would result in the same number, the number 1
act as the neutral multiplier in multiplication. Example: 1 * 1 = 1
.
This is why we use a starting value of 1
for multiplication and a starting value of 0
for addition.
Example:
const numbers = [1, 2, 5];
const finalResult = numbers.reduce((accumulator, current) => {
return accumulator * current;
}, 1);
Detailed Explanation:
The first time the callback is executed, accumulator is set to 1 (due to the initial value) and current is set to the first element of the array. Consequently, accumulator
is 1 and current
is 1.
Then we output accumulator * current
, which is 1 * 1 = 1. The updated value of the accumulator
is 1.
The callback executes a second time, with current = 2
(the second item in the array) and accumulator = 1
. We return accumulator * current
, which is 1 * 2 = 2
. The accumulator
is 2.
The callback executes a third time, with current = 5
(the third item in the array) and accumulator = 2
. We return accumulator * current
, which is 2 * 5 = 10
. The accumulator
is 10.
The result of the .reduce()
is 10 which is stored in the variable finalResult.
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