From: "marcandre (Marc-Andre Lafortune)" <ruby-core@...>
Date: 2013-09-12T02:41:17+09:00
Subject: [ruby-core:57145] [ruby-trunk - Feature #8895] Destructuring	Assignment for Hash


Issue #8895 has been updated by marcandre (Marc-Andre Lafortune).


I suggested something similar in  [ruby-core:41772].
Here is a summary from my similar suggestion made in [ruby-core:41772]:

  {key: 'default', other_key:, **other_options} = {other_key: 42, foo: 'bar'}
  key # => 'default'
  other_key # => 42
  other_options # => {foo: 'bar'}

You'll note that it doesn't give the possibility to map the key to a different variable. Indeed, I don't think that it would be useful and I would rather encourage rubyists to use meanginful option and variable names. It also makes very similar to the way we declare keyword arguments for methods, so no additional learning curve. Your proposal is quite different.

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Feature #8895: Destructuring Assignment for Hash
https://bugs.ruby-lang.org/issues/8895#change-41758

Author: chendo (Jack Chen)
Status: Open
Priority: Normal
Assignee: 
Category: 
Target version: 


=begin
Given Ruby already supports destructuring assignment with Array (a, b = [1, 2]), I propose destructuring assignments for Hash.

== Basic example

  params = {name: "John Smith", age: 42}
  {name: name, age: age} = params

  # name == "John Smith"
  # age == 42

This would replace a common pattern of assigning hash values to local variables to work with.

== General syntax

  { <key-expr> => <variable_name>, ��� } = <object that responds to #[]>

  # Symbols
  { foo: bar } = { foo: "bar" }
  bar == "bar"

  # Potential shorthand
  { foo } = { foo: "bar" }
  foo == "bar"

== Use cases:

  # MatchData
  { username: username, age: age } = "user:jsmith age:42".match(/user:(?<username>\w+) age:(?<age>\d+)/)
  username == "jsmith"
  age == "42"

== Edge cases

  # Variable being assigned to more than once should use the last one
  { foo: var, bar: var } = {foo: 1, bar: 2}
  var == 2

Thoughts?
=end



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