Javascript Reduce Throwing Error For More Than Two Items

I have this array: const arr = [ { someProp: [{ amount: 10 }]}, { someProp: [{ amount: 12 }]}, ]; and then this reduce fn: const sum = arr.reduce((prev, curr) => prev.someP

Solution 1:

Actually you need only to return the accumulator + the new value.

You want to begin with 0 i guess so you need to add 0 as second parameter

const arr =
    [
      { someProp: [{ amount: 10 }]},
      { someProp: [{ amount: 12 }]},
      { someProp: [{ amount: 12 }]},
    ];
    
const sum = arr.reduce((acc, item) => item.someProp[0].amount + acc, 0);

console.log(sum);

If you want it more spicy here with destructuring

const arr =
    [
      { someProp: [{ amount: 10 }]},
      { someProp: [{ amount: 12 }]},
      { someProp: [{ amount: 12 }]},
    ];
    
const sum = arr.reduce((acc, { someProp: [{ amount }] }) => amount + acc, 0);

console.log(sum);

Solution 2:

Following @Pointy answer, this is how you can get it working with any number of elements:

const sum = arr.reduce((prev, curr) => prev + curr.someProp[0].amount, 0);

prev is always the previously returned element, so you can safely use it as a number (and since you're just summing it, you can use 0 as default, so the first time reduce is invoked the value of prev will be 0)

Solution 3:

You're returning a number. The value of prev is always the prior return value from the callback, not the previous element. Thus on the third iteration you're trying to use that number as if it were an object reference.

Add a second parameter (0) to the .reduce() call, and change the function to treat prev as the running total (a simple number):

const sum = arr.reduce((prev, curr) => prev + curr.someProp[0].amount, 0);

It works as-is when there are two elements because without that second parameter, the first iteration will see element 0 as prev and element 1 as curr. That behavior works fine when you've got an array of numbers and you just want to perform a computation between them, but in this case you need that initial value explicitly there as the second argument.