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Ruby method Array#<< not updating the array in hash
Strange ruby behavior when using Hash.new([])
I've been doing Koans which is great, and as I go along I find no major trouble, but I stumbled upon this, and can't make any sense out of it:
def test_default_value_is_the_same_object
hash = Hash.new([])
hash[:one] << "uno"
hash[:two] << "dos"
assert_equal ["uno", "dos"], hash[:one] # But I only put "uno" for this key!
assert_equal ["uno", "dos"], hash[:two] # But I only put "dos" for this key!
assert_equal ["uno", "dos"], hash[:three] # I didn't shove anything for :three!
assert_equal true, hash[:one].object_id == hash[:two].object_id
end
All the tests are passing (I just looked at the error which helped me guess the right assertions to write).
The last assert, ok, they both were not initialized so their values have got to have the same object ID since they both take the default.
I don't understand why the default value was altered, I'm not even entirely sure that's what happened.
I tried it out in IRB, thinking maybe some tampering on Hash/Array was done to make me crazy, but I get the same result.
I first thought hash[:one] << "uno" would imply hash to become { one: ["uno] }, but it remains { }.
Although I'm guessing << only calls push, and new keys are only added when you use the = sign
Please tell me what I missed.
EDIT: I'm using Ruby 1.9.3
When you use the default argument for a Hash, the same object is used for all keys that have not been explicitly set. This means that only one array is being used here, the one you passed into Hash.new. See below for evidence of that.
>> h = Hash.new([])
=> {}
>> h[:foo] << :bar
=> [:bar]
>> h[:bar] << :baz
=> [:bar, :baz]
>> h[:foo].object_id
=> 2177177000
>> h[:bar].object_id
=> 2177177000
The weird thing is that as you found, if you inspect the hash, you'll find that it is empty! This is because only the default object has been modified, no keys have yet been assigned.
Fortunately, there is another way to do default values for hashes. You can provide a default block instead:
>> h = Hash.new { |h,k| h[k] = [] }
=> {}
>> h[:foo] << :bar
=> [:bar]
>> h[:bar] << :baz
=> [:baz]
>> h[:foo].object_id
=> 2176949560
>> h[:bar].object_id
=> 2176921940
When you use this approach, the block gets executed every time an unassigned key is used, and it is provided the hash itself and the key as an argument. By assigning the default value within the block, you can be sure that a new object will get created for each distinct key, and that the assignment will happen automatically. This is the idiomatic way of creating a "Hash of Arrays" in Ruby, and is generally safer to use than the default argument approach.
That said, if you're working with immutable values (like numbers), doing something like Hash.new(0) is safe, as you'll only change those values by re-assignment. But because I prefer to keep fewer concepts in my head, I pretty much use the block form exclusively.
When you do
h = Hash.new(0)
h[:foo] += 1
you are directly modifying h. h[:foo] += 1 is the same as h[:foo] = h[:foo]+1. h[:foo] is being assigned 0+1.
When you do
h = Hash.new([])
h[:foo] << :bar
you are modifying h[:foo] which is [], which is the default value of h but is not a value to any key of h. After that [] becomes [:bar], the default value of h becomes [:bar], but that is not the value for h[:foo].
Related
So let's say I do the following:
lph = Hash.new([]) #=> {}
lph["passed"] << "LCEOT" #=> ["LCEOT"]
lph #=> {} <-- Expected that to have been {"passed" => ["LCEOT"]}
lph["passed"] #=> ["LCEOT"]
lph["passed"] = lph["passed"] << "HJKL"
lph #=> {"passed"=>["LCEOT", "HJKL"]}
I'm surprised by this. A couple questions:
Why does it not get set until I push the second string on to the array? What is happening in the background?
What is the more idiomatic ruby way to essentially say. I have a hash, a key, and a value I want to to end up in the array associated with the key. How do I push the value in an array associated with a key into a hash the first time. In all future uses of the key, I just want to addd to the array.
Read the Ruby Hash.new documentation carefully - "if this hash is subsequently accessed by a key that doesn’t correspond to a hash entry, the value returned depends on the style of new used to create the hash".
new(obj) → new_hash
...If obj is specified, this single object will be used for all default values.
In your example you attempt to push something onto the value associated with a key which does not exist, so you end up mutating the same anonymous array you used to construct the hash initially.
the_array = []
h = Hash.new(the_array)
h['foo'] << 1 # => [1]
# Since the key 'foo' was not found
# ... the default value (the_array) is returned
# ... and 1 is pushed onto it (hence [1]).
the_array # => [1]
h # {} since the key 'foo' still has no value.
You probably want to use the block form:
new { |hash, key| block } → new_hash
...If a block is specified, it will be called with the hash object and the key, and should return the default value. It is the block’s responsibility to store the value in the hash if required.
For example:
h = Hash.new { |hash, key| hash[key] = [] } # Assign a new array as default for missing keys.
h['foo'] << 1 # => [1]
h['foo'] << 2 # => [1, 2]
h['bar'] << 3 # => [3]
h # => { 'foo' => [1, 2], 'bar' => [3] }
Why does it not get set until I push the second string on to the array?
In short; because you don't set anything in the hash until the point, where you also add the second string to the array.
What is happening in the background?
To see what's happening in the background, let's take this one line at a time:
lph = Hash.new([]) #=> {}
This creates an empty hash, configured to return the [] object whenever a non-existing key is accessed.
lph["passed"] << "LCEOT" #=> ["LCEOT"]
This can be written as
value = lph["passed"] #=> []
value << "LCEOT" #=> ["LCEOT"]
We see that lph["passed"] returns [] as expected, and we then proceed to append "LCEOT" to [].
lph #=> {}
lph is still an empty Hash. At no point have we added anything to the Hash. We have added something to its default value, but that doesn't change lph itself.
lph["passed"] #=> ["LCEOT"]
This is where it gets interesting. Remember above when we did value << ["LCEOT"]. That actually changed the default value that lph returns when a key isn't found. The default value is no longer [], but has become ["LCEOT"]. That new default value is returned here.
lph["passed"] = lph["passed"] << "HJKL"
This is our first change to lph. And what we actually assign to lph["passed"] is the default value (because "passed" is still a non-existing key in lph) with "HJKL" appended. Before this, the default value was ["LCEOT"], after this it is ["LCEOT", "HJKL"].
In other words lph["passed"] << "HJKL" returns ["LCEOT", "HJKL"] which is then assigned to lph["passed"].
What is the more idiomatic Ruby way
Using <<=:
>> lph = Hash.new { [] }
=> {}
>> lph["passed"] <<= "LCEOT"
=> ["LCEOT"]
>> lph
=> {"passed"=>["LCEOT"]}
Also note the change in how the Hash is initialized, using a block instead of a verbatim array. This ensures a new, blank array is created and returned whenever a new key is accessed, as opposed to the same array being used every time.
There's a nice idiom for adding to lists stored in a hash table:
(hash[key] ||= []) << new_value
Now, suppose I write a derivative hash class, like the ones found in Hashie, which does a deep-convert of any hash I store in it. Then what I store will not be the same object I passed to the = operator; Hash may be converted to Mash or Clash, and arrays may be copied.
Here's the problem. Ruby apparently returns, from the var= method, the value passed in, not the value that's stored. It doesn't matter what the var= method returns. The code below demonstrates this:
class C
attr_reader :foo
def foo=(value)
#foo = (value.is_a? Array) ? (value.clone) : value
end
end
c=C.new
puts "assignment: #{(c.foo ||= []) << 5}"
puts "c.foo is #{c.foo}"
puts "assignment: #{(c.foo ||= []) << 6}"
puts "c.foo is #{c.foo}"
output is
assignment: [5]
c.foo is []
assignment: [6]
c.foo is [6]
When I posted this as a bug to Hashie, Danielle Sucher explained what was happening and pointed out that "foo.send :bar=, 1" returns the value returned by the bar= method. (Hat tip for the research!) So I guess I could do:
c=C.new
puts "clunky assignment: #{(c.foo || c.send(:foo=, [])) << 5}"
puts "c.foo is #{c.foo}"
puts "assignment: #{(c.foo || c.send(:foo=, [])) << 6}"
puts "c.foo is #{c.foo}"
which prints
clunky assignment: [5]
c.foo is [5]
assignment: [5, 6]
c.foo is [5, 6]
Is there any more elegant way to do this?
Assignments evaluate to the value that is being assigned. Period.
In some other languages, assignments are statements, so they don't evaluate to anything. Those are really the only two sensible choices. Either don't evaluate to anything, or evaluate to the value being assigned. Everything else would be too surprising.
Since Ruby doesn't have statements, there is really only one choice.
The only "workaround" for this is: don't use assignment.
c.foo ||= []
c.foo << 5
Using two lines of code isn't the end of the world, and it's easier on the eyes.
The prettiest way to do this is to use default value for hash:
# h = Hash.new { [] }
h = Hash.new { |h,k| h[k] = [] }
But be ware that you cant use Hash.new([]) and then << because of way how Ruby store variables:
h = Hash.new([])
h[:a] # => []
h[:b] # => []
h[:a] << 10
h[:b] # => [10] O.o
it's caused by that Ruby store variables by reference, so as we created only one array instance, ad set it as default value then it will be shared between all hash cells (unless it will be overwrite, i.e. by h[:a] += [10]).
It is solved by using constructor with block (doc) Hash.new { [] }. With this each time when new key is created block is called and each value is different array.
EDIT: Fixed error that #Uri Agassi is writing about.
I'm going through Ruby Koans, and I hit #41 which I believe is this:
def test_default_value_is_the_same_object
hash = Hash.new([])
hash[:one] << "uno"
hash[:two] << "dos"
assert_equal ["uno","dos"], hash[:one]
assert_equal ["uno","dos"], hash[:two]
assert_equal ["uno","dos"], hash[:three]
assert_equal true, hash[:one].object_id == hash[:two].object_id
end
It could not understand the behavior so I Googled it and found Strange ruby behavior when using Hash default value, e.g. Hash.new([]) that answered the question nicely.
So I understand how that works, my question is, why does a default value such as an integer that gets incremented not get changed during use? For example:
puts "Text please: "
text = gets.chomp
words = text.split(" ")
frequencies = Hash.new(0)
words.each { |word| frequencies[word] += 1 }
This will take user input and count the number of times each word is used, it works because the default value of 0 is always used.
I have a feeling it has to do with the << operator but I'd love an explanation.
The other answers seem to indicate that the difference in behavior is due to Integers being immutable and Arrays being mutable. But that is misleading. The difference is not that the creator of Ruby decided to make one immutable and the other mutable. The difference is that you, the programmer decided to mutate one but not the other.
The question is not whether Arrays are mutable, the question is whether you mutate it.
You can get both the behaviors you see above, just by using Arrays. Observe:
One default Array with mutation
hsh = Hash.new([])
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => ['one', 'two']
# Because we mutated the default value, nonexistent keys return the changed value
hsh
# => {}
# But we never mutated the hash itself, therefore it is still empty!
One default Array without mutation
hsh = Hash.new([])
hsh[:one] += ['one']
hsh[:two] += ['two']
# This is syntactic sugar for hsh[:two] = hsh[:two] + ['two']
hsh[:nonexistant]
# => []
# We didn't mutate the default value, it is still an empty array
hsh
# => { :one => ['one'], :two => ['two'] }
# This time, we *did* mutate the hash.
A new, different Array every time with mutation
hsh = Hash.new { [] }
# This time, instead of a default *value*, we use a default *block*
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => []
# We *did* mutate the default value, but it was a fresh one every time.
hsh
# => {}
# But we never mutated the hash itself, therefore it is still empty!
hsh = Hash.new {|hsh, key| hsh[key] = [] }
# This time, instead of a default *value*, we use a default *block*
# And the block not only *returns* the default value, it also *assigns* it
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => []
# We *did* mutate the default value, but it was a fresh one every time.
hsh
# => { :one => ['one'], :two => ['two'], :nonexistent => [] }
It is because Array in Ruby is mutable object, so you can change it internal state, but Fixnum isn't mutable. So when you increment value using += internally it get that (assume that i is our reference to Fixnum object):
get object referenced by i
get it internal value (lets name it raw_tmp)
create new object that internal value is raw_tmp + 1
assign reference to created object to i
So as you can see, we created new object, and i reference now to something different than at the beginning.
In the other hand, when we use Array#<< it works that way:
get object referenced by arr
to it's internal state append given element
So as you can see it is much simpler, but it can cause some bugs. One of them you have in your question, another one is thread race when booth are trying simultaneously append 2 or more elements. Sometimes you can end with only some of them and with thrashes in memory, when you use += on arrays too, you will get rid of both of these problems (or at least minimise impact).
From the doc, setting a default value has the following behaviour:
Returns the default value, the value that would be returned by hsh if key did not exist in hsh. See also Hash::new and Hash#default=.
Therefore, every time frequencies[word] is not set, the value for that individual key is set to 0.
The reason for the discrepancy between the two code blocks is that arrays are mutable in Ruby, while integers are not.
I was going through Ruby Koans tutorial series, when I came upon this in about_hashes.rb:
def test_default_value_is_the_same_object
hash = Hash.new([])
hash[:one] << "uno"
hash[:two] << "dos"
assert_equal ["uno", "dos"], hash[:one]
assert_equal ["uno", "dos"], hash[:two]
assert_equal ["uno", "dos"], hash[:three]
assert_equal true, hash[:one].object_id == hash[:two].object_id
end
The values in assert_equals, is actually what the tutorial expected. But I couldn't understand how there is a difference between using << operator and = operator?
My expectation was that:
hash[:one] would be ["uno"]
hash[:two] would be ["dos"]
hash[:three] would be []
Can someone please explain why my expectation was wrong?
You have mixed up the way this works a bit. First off, a Hash doesn't have a << method, that method in your example exists on the array.
The reason your code is not erroring is because you are passing a default value to your hash via the constructor. http://ruby-doc.org/core-1.9.3/Hash.html#method-c-new
hash = Hash.new([])
This means that if a key does not exist, then it will return an array. If you run the following code:
hash = {}
hash[:one] << "uno"
Then you will get an undefined method error.
So in your example, what is actually happening is:
hash = Hash.new([])
hash[:one] << "uno" #hash[:one] does not exist so return an array and push "uno"
hash[:two] << "dos" #hash[:two] does not exist, so return the array ["uno"] and push "dos"
The reason it does not return an array with one element each time as you may expect, is because it stores a reference to the value that you pass through to the constructor. Meaning that each time an element is pushed, it modifies the initial array.
When you're doing hash = Hash.new([]) you are creating a Hash whose default value is the exact same Array instance for all keys. So whenever you are accessing a key that doesn't exist, you get back the very same Array.
h = Hash.new([])
h[:foo].object_id # => 12215540
h[:bar].object_id # => 12215540
If you want one array per key, you have to use the block syntax of Hash.new:
h = Hash.new { |h, k| h[k] = [] }
h[:foo].object_id # => 7791280
h[:bar].object_id # => 7790760
Edit: Also see what Gazler has to say with regard to the #<< method and on what object you are actually calling it.
This question already has answers here:
Ruby Style: How to check whether a nested hash element exists
(16 answers)
How to avoid NoMethodError for nil elements when accessing nested hashes? [duplicate]
(4 answers)
Closed 7 years ago.
I'm working a little utility written in ruby that makes extensive use of nested hashes. Currently, I'm checking access to nested hash elements as follows:
structure = { :a => { :b => 'foo' }}
# I want structure[:a][:b]
value = nil
if structure.has_key?(:a) && structure[:a].has_key?(:b) then
value = structure[:a][:b]
end
Is there a better way to do this? I'd like to be able to say:
value = structure[:a][:b]
And get nil if :a is not a key in structure, etc.
Traditionally, you really had to do something like this:
structure[:a] && structure[:a][:b]
However, Ruby 2.3 added a method Hash#dig that makes this way more graceful:
structure.dig :a, :b # nil if it misses anywhere along the way
There is a gem called ruby_dig that will back-patch this for you.
Hash and Array have a method called dig.
value = structure.dig(:a, :b)
It returns nil if the key is missing at any level.
If you are using a version of Ruby older than 2.3, you can install a gem such as ruby_dig or hash_dig_and_collect, or implement this functionality yourself:
module RubyDig
def dig(key, *rest)
if value = (self[key] rescue nil)
if rest.empty?
value
elsif value.respond_to?(:dig)
value.dig(*rest)
end
end
end
end
if RUBY_VERSION < '2.3'
Array.send(:include, RubyDig)
Hash.send(:include, RubyDig)
end
The way I usually do this these days is:
h = Hash.new { |h,k| h[k] = {} }
This will give you a hash that creates a new hash as the entry for a missing key, but returns nil for the second level of key:
h['foo'] -> {}
h['foo']['bar'] -> nil
You can nest this to add multiple layers that can be addressed this way:
h = Hash.new { |h, k| h[k] = Hash.new { |hh, kk| hh[kk] = {} } }
h['bar'] -> {}
h['tar']['zar'] -> {}
h['scar']['far']['mar'] -> nil
You can also chain indefinitely by using the default_proc method:
h = Hash.new { |h, k| h[k] = Hash.new(&h.default_proc) }
h['bar'] -> {}
h['tar']['star']['par'] -> {}
The above code creates a hash whose default proc creates a new Hash with the same default proc. So, a hash created as a default value when a lookup for an unseen key occurs will have the same default behavior.
EDIT: More details
Ruby hashes allow you to control how default values are created when a lookup occurs for a new key. When specified, this behavior is encapsulated as a Proc object and is reachable via the default_proc and default_proc= methods. The default proc can also be specified by passing a block to Hash.new.
Let's break this code down a little. This is not idiomatic ruby, but it's easier to break it out into multiple lines:
1. recursive_hash = Hash.new do |h, k|
2. h[k] = Hash.new(&h.default_proc)
3. end
Line 1 declares a variable recursive_hash to be a new Hash and begins a block to be recursive_hash's default_proc. The block is passed two objects: h, which is the Hash instance the key lookup is being performed on, and k, the key being looked up.
Line 2 sets the default value in the hash to a new Hash instance. The default behavior for this hash is supplied by passing a Proc created from the default_proc of the hash the lookup is occurring in; ie, the default proc the block itself is defining.
Here's an example from an IRB session:
irb(main):011:0> recursive_hash = Hash.new do |h,k|
irb(main):012:1* h[k] = Hash.new(&h.default_proc)
irb(main):013:1> end
=> {}
irb(main):014:0> recursive_hash[:foo]
=> {}
irb(main):015:0> recursive_hash
=> {:foo=>{}}
When the hash at recursive_hash[:foo] was created, its default_proc was supplied by recursive_hash's default_proc. This has two effects:
The default behavior for recursive_hash[:foo] is the same as recursive_hash.
The default behavior for hashes created by recursive_hash[:foo]'s default_proc will be the same as recursive_hash.
So, continuing in IRB, we get the following:
irb(main):016:0> recursive_hash[:foo][:bar]
=> {}
irb(main):017:0> recursive_hash
=> {:foo=>{:bar=>{}}}
irb(main):018:0> recursive_hash[:foo][:bar][:zap]
=> {}
irb(main):019:0> recursive_hash
=> {:foo=>{:bar=>{:zap=>{}}}}
I made rubygem for this. Try vine.
Install:
gem install vine
Usage:
hash.access("a.b.c")
I think one of the most readable solutions is using Hashie:
require 'hashie'
myhash = Hashie::Mash.new({foo: {bar: "blah" }})
myhash.foo.bar
=> "blah"
myhash.foo?
=> true
# use "underscore dot" for multi-level testing
myhash.foo_.bar?
=> true
myhash.foo_.huh_.what?
=> false
value = structure[:a][:b] rescue nil
Solution 1
I suggested this in my question before:
class NilClass; def to_hash; {} end end
Hash#to_hash is already defined, and returns self. Then you can do:
value = structure[:a].to_hash[:b]
The to_hash ensures that you get an empty hash when the previous key search fails.
Solution2
This solution is similar in spirit to mu is too short's answer in that it uses a subclass, but still somewhat different. In case there is no value for a certain key, it does not use a default value, but rather creates a value of empty hash, so that it does not have the problem of confusion in assigment that DigitalRoss's answer has, as was pointed out by mu is too short.
class NilFreeHash < Hash
def [] key; key?(key) ? super(key) : self[key] = NilFreeHash.new end
end
structure = NilFreeHash.new
structure[:a][:b] = 3
p strucrture[:a][:b] # => 3
It departs from the specification given in the question, though. When an undefined key is given, it will return an empty hash instread of nil.
p structure[:c] # => {}
If you build an instance of this NilFreeHash from the beginning and assign the key-values, it will work, but if you want to convert a hash into an instance of this class, that may be a problem.
You could just build a Hash subclass with an extra variadic method for digging all the way down with appropriate checks along the way. Something like this (with a better name of course):
class Thing < Hash
def find(*path)
path.inject(self) { |h, x| return nil if(!h.is_a?(Thing) || h[x].nil?); h[x] }
end
end
Then just use Things instead of hashes:
>> x = Thing.new
=> {}
>> x[:a] = Thing.new
=> {}
>> x[:a][:b] = 'k'
=> "k"
>> x.find(:a)
=> {:b=>"k"}
>> x.find(:a, :b)
=> "k"
>> x.find(:a, :b, :c)
=> nil
>> x.find(:a, :c, :d)
=> nil
This monkey patch function for Hash should be easiest (at least for me). It also doesn't alter structure i.e. changing nil's to {}. It would still also apply even if you're reading a tree from a raw source e.g. JSON. It also doesn't need to produce empty hash objects as it goes or parse a string. rescue nil was actually a good easy solution for me as I'm brave enough for such a low risk but I find it to essentially have a drawback with performance.
class ::Hash
def recurse(*keys)
v = self[keys.shift]
while keys.length > 0
return nil if not v.is_a? Hash
v = v[keys.shift]
end
v
end
end
Example:
> structure = { :a => { :b => 'foo' }}
=> {:a=>{:b=>"foo"}}
> structure.recurse(:a, :b)
=> "foo"
> structure.recurse(:a, :x)
=> nil
What's also good is that you can play around saved arrays with it:
> keys = [:a, :b]
=> [:a, :b]
> structure.recurse(*keys)
=> "foo"
> structure.recurse(*keys, :x1, :x2)
=> nil
The XKeys gem will read and auto-vivify-on-write nested hashes (::Hash) or hashes and arrays (::Auto, based on the key/index type) with a simple, clear, readable, and compact syntax by enhancing #[] and #[]=. The sentinel symbol :[] will push onto the end of an array.
require 'xkeys'
structure = {}.extend XKeys::Hash
structure[:a, :b] # nil
structure[:a, :b, :else => 0] # 0 (contextual default)
structure[:a] # nil, even after above
structure[:a, :b] = 'foo'
structure[:a, :b] # foo
You can use the andand gem, but I'm becoming more and more wary of it:
>> structure = { :a => { :b => 'foo' }} #=> {:a=>{:b=>"foo"}}
>> require 'andand' #=> true
>> structure[:a].andand[:b] #=> "foo"
>> structure[:c].andand[:b] #=> nil
There is the cute but wrong way to do this. Which is to monkey-patch NilClass to add a [] method that returns nil. I say it is the wrong approach because you have no idea what other software may have made a different version, or what behavior change in a future version of Ruby can be broken by this.
A better approach is to create a new object that works a lot like nil but supports this behavior. Make this new object the default return of your hashes. And then it will just work.
Alternately you can create a simple "nested lookup" function that you pass the hash and the keys to, which traverses the hashes in order, breaking out when it can.
I would personally prefer one of the latter two approaches. Though I think it would be cute if the first was integrated into the Ruby language. (But monkey-patching is a bad idea. Don't do that. Particularly not to demonstrate what a cool hacker you are.)
Not that I would do it, but you can Monkeypatch in NilClass#[]:
> structure = { :a => { :b => 'foo' }}
#=> {:a=>{:b=>"foo"}}
> structure[:x][:y]
NoMethodError: undefined method `[]' for nil:NilClass
from (irb):2
from C:/Ruby/bin/irb:12:in `<main>'
> class NilClass; def [](*a); end; end
#=> nil
> structure[:x][:y]
#=> nil
> structure[:a][:y]
#=> nil
> structure[:a][:b]
#=> "foo"
Go with #DigitalRoss's answer. Yes, it's more typing, but that's because it's safer.
In my case, I needed a two-dimensional matrix where each cell is a list of items.
I found this technique which seems to work. It might work for the OP:
$all = Hash.new()
def $all.[](k)
v = fetch(k, nil)
return v if v
h = Hash.new()
def h.[](k2)
v = fetch(k2, nil)
return v if v
list = Array.new()
store(k2, list)
return list
end
store(k, h)
return h
end
$all['g1-a']['g2-a'] << '1'
$all['g1-a']['g2-a'] << '2'
$all['g1-a']['g2-a'] << '3'
$all['g1-a']['g2-b'] << '4'
$all['g1-b']['g2-a'] << '5'
$all['g1-b']['g2-c'] << '6'
$all.keys.each do |group1|
$all[group1].keys.each do |group2|
$all[group1][group2].each do |item|
puts "#{group1} #{group2} #{item}"
end
end
end
The output is:
$ ruby -v && ruby t.rb
ruby 1.9.2p0 (2010-08-18 revision 29036) [x86_64-linux]
g1-a g2-a 1
g1-a g2-a 2
g1-a g2-a 3
g1-a g2-b 4
g1-b g2-a 5
g1-b g2-c 6
I am currently trying out this:
# --------------------------------------------------------------------
# System so that we chain methods together without worrying about nil
# values (a la Objective-c).
# Example:
# params[:foo].try?[:bar]
#
class Object
# Returns self, unless NilClass (see below)
def try?
self
end
end
class NilClass
class MethodMissingSink
include Singleton
def method_missing(meth, *args, &block)
end
end
def try?
MethodMissingSink.instance
end
end
I know the arguments against try, but it is useful when looking into things, like say, params.