From: ko1@...
Date: 2014-07-10T07:11:36+00:00
Subject: [ruby-core:63622] [ruby-trunk - Feature #8895] Destructuring	Assignment for Hash

Issue #8895 has been updated by Koichi Sasada.


+1 for this proposal.

I feel it is fine for me:

```ruby
  k1: 1, k2: 2 = h
  kr1:, kr2: = h

  #=> same as
  k1 = h.fetch(:k1, 1)
  k2 = h.fetch(:k2, 2)
  kr1 = h.fetch(:k1)
  kr2 = h.fetch(:k2)

  # mixed
  k1: 1, k2: 2, kr1:, kr2: = h

  # compile to
  k1 = h.fetch(:k1, 1)
  k2 = h.fetch(:k2, 2)
  kr1 = h.fetch(:k1)
  kr2 = h.fetch(:k2)
```

Problem is what happen when `h' is not a hash object (and doesn't have to_hash method).
Just ignore is one option (what ary assignment do, like "a, b = 1 #=> 1, nil").

> a, *b, c:, d: 'd', **e = [1, {c: 2}]

It should be:

```ruby
a, (c:, d: 'd', **e) = [1, {c:2}]
```


----------------------------------------
Feature #8895: Destructuring Assignment for Hash
https://bugs.ruby-lang.org/issues/8895#change-47675

* Author: Jack Chen
* Status: Open
* Priority: Normal
* Assignee: 
* Category: 
* Target version: 
----------------------------------------

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

Basic example
-------------

```ruby
  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
--------------

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

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

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

Use cases
---------

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

Edge cases
----------

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

Thoughts?





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