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What does this mean in Ruby language?

开发者 https://www.devze.com 2023-01-13 20:18 出处:网络
Run the following code, a = [1, 2, 3, 4, 5] head, *tail = a p head p tail You will get the result 1 [2, 3, 4, 开发者_开发技巧5]

Run the following code,

a = [1, 2, 3, 4, 5]
head, *tail = a
p head
p tail

You will get the result

1
[2, 3, 4, 开发者_开发技巧5]

Who can help me to explain the statement head,*tail = a, Thanks!


head, *tail = a means to assign the first element of the array a to head, and assign the rest of the elements to tail.

*, sometimes called the "splat operator," does a number of things with arrays. When it's on the left side of an assignment operator (=), as in your example, it just means "take everything left over."

If you omitted the splat in that code, it would do this instead:

head, tail = [1, 2, 3, 4, 5]
p head # => 1
p tail # => 2

But when you add the splat to tail it means "Everything that didn't get assigned to the previous variables (head), assign to tail."


First, it is a parallel assignment. In ruby you can write

a,b = 1,2

and a will be 1 and b will be 2. You can also use

a,b = b,a

to swap values (without the typical temp-variable needed in other languages).

The star * is the pack/unpack operator. Writing

a,b = [1,2,3]

would assign 1 to a and 2 to b. By using the star, the values 2,3 are packed into an array and assigned to b:

a,*b = [1,2,3]


I don't know Ruby at all, but my guess is that the statement is splitting the list a into a head (first element) and the rest (another list), assigning the new values to the variables head and tail.

This mechanism is usually referred (at least in Erlang) as pattern matching.

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