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Start your free trialvamosrope14
14,932 PointsDid we need to specify the default value of the accumulator to be equal to "[]"?
.
1 Answer
Steven Parker
231,184 PointsIf 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.
Will Albertsen
13,155 PointsWill Albertsen
13,155 PointsSo 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"]
Steven Parker
231,184 PointsSteven Parker
231,184 PointsSeparating the strings into individual characters is not the objective of the exercise.
Tyler McDonald
Full Stack JavaScript Techdegree Graduate 16,700 PointsTyler McDonald
Full Stack JavaScript Techdegree Graduate 16,700 PointsHello 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
Steven Parker
231,184 PointsSteven Parker
231,184 PointsYou'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.