Join arrays

Explanation

Let's detail step-by-step what occurs in the following line of code

["a", *["b", "c"], "d"] * ?: # => "a:b:c:d"

Array Destructuring

*["b", "c"] concatenates the sub-array elements into the main array.

So after this operation, the array looks like this:

a = ["a", *["b", "c"], "d"]

a # => ["a", "b", "c", "d"]

This idiom is particularly used in CSV generation when the number of columns can only be known at runtime.

book = Book.preload(:authors).find(123)

row = [book.title, book.short_description, *book.authors.pluck(:fullname)]

row # => ["Hello", "A book about Hello", "Author 1", "Author 2", "Author 3"]

Single Character String

?: is a shorthand to create a single character string.

So

["a", "b", "c", "d"] * ?:

is equivalent to

["a", "b", "c", "d"] * ':'

Array#*

When Array#* takes a string as an argument, it acts like Array#join(sep).

Here

["a", "b", "c", "d"] * ':'

is equivalent to

["a", "b", "c", "d"].join(':')

So the return value of ["a", "b", "c", "d"] * ?: is "a:b:c:d".

Voilà!