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# Did we need to specify the default value of the accumulator to be equal to "[]"?

For example, we write [...arr...curr],[]---> the initial value of the accumulator is [], but wouldn't it be okay to omit the initial value so that by default, the initial value is one the inner rays? Wouldn't that be fine?

Because at the end of the day you just want one array all values separated by commas.

If you don't supply an initial value, the operation will start with the first element in the accumulator, so the accumulator would not be an array.

So how come the code still works normally even if an empty array is not set as the initial value? Since the first value of the accumulator would be the string 'The Day the Earth Stood Still', would calling the spread operator on it not get you the following result?

```["T", "h", "e", " ", "D", "a", "y", " ", "t", "h", "e", " ", "E", "a", "r", "t", "h", " ", "S", "t", "o", "o", "d", " ", "S", "t", "i", "l", "l"]
```

Separating the strings into individual characters is not the objective of the exercise.

Hello from 2022! I am wondering the same thing. I ran the code both ways, and they both return an array of the string values, with or without the initial value. Am I missing something?

With initial value:

```const flatMovies = movies.reduce((arr, innerMovies) => [...arr, ...innerMovies], []);
```

vs.

Without initial value:

```const flatMovies = movies.reduce((arr, innerMovies) => [...arr, ...innerMovies]);
```

Both return an array

```[
'The Day the Earth Stood Still',
'Superman',
'Ghostbusters',
'Finding Dory',
'Jaws',
'On the Waterfront'
]
```

You're right, since the individual elements are arrays. I'm wondering if the example was different in 2019, since leaving off the initial value also doesn't produce the result shown by Will Albertsen now, either.