So i discovered this ruby behaviour, which kept me going crazy for over an hour. When I pass a hash to a function which has a default value for hash AND a keyword argument, it seems like the reference doesn't get passed correctly. As soon as I take away the default value OR the keyword argument, the function behaves as expected. Am I missing some obvious ruby rule here?
def change_hash(h={}, rand: om)
h['hey'] = true
end
k = {}
change_hash(k)
k
#=> {}
It works fine as soon as I take out the default or the keyword arg.
def change_hash(h, rand: om)
h['hey'] = true
end
k = {}
change_hash(k)
k
#=> {'hey' => true}
def change_hash(h={})
h['hey'] = true
end
k = {}
change_hash(k)
k
#=> {'hey' => true}
EDIT
Thanks for your answers. Most of you pointed out that ruby parses the hash as a keyword argument in some cases. However, I am talking about the case when a hash has string keys. When I pass the hash, it seems like the value that gets passed is correct. But modifying the hash inside the function doesn't modify the original hash.
def change_hash(hash={}, another_arg: 300)
puts "another_arg: #{another_arg}"
puts "hash: #{hash}"
hash['hey'] = 3
end
my_hash = {"o" => 3}
change_hash(my_hash)
puts my_hash
Prints out
another_arg: 300
hash: {"o"=>3}
{"o"=>3}
TL;DR ruby allows passing hash as a keyword argument as well as “expanded inplace hash.” Since change_hash(rand: :om) must be routed to keyword argument, so should change_hash({rand: :om}) and, hence, change_hash({}).
Since ruby allows default arguments in any position, the parser takes care of default arguments in the first place. That means, that the default arguments are greedy and the most amount of defaults will take a place.
On the other hand, since ruby lacks pattern-matching feature for function clauses, parsing the given argument to decide whether it should be passed as double-splat or not would lead to huge performance penalties. Since the call with an explicit keyword argument (change_hash(rand: :om)) should definitely pass :om to keyword argument, and we are allowed to pass an explicit hash {rand: :om} as a keyword argument, Ruby has nothing to do but to accept any hash as a keyword argument.
Ruby will split the single hash argument between hash and rand:
k = {"a" => 42, rand: 42}
def change_hash(h={}, rand: :om)
h[:foo] = 42
puts h.inspect
end
change_hash(k);
puts k.inspect
#⇒ {"a"=>42, :foo=>42}
#⇒ {"a"=>42, :rand=>42}
That split feature requires the argument being cloned before passing. That is why the original hash is not being modified.
This is particularly tricky case in Ruby indeed.
In your example you have optional argument which is a hash and you have an optional keyword argument at the same time. In this situation if you pass only one hash, Ruby interprets it as a hash which contains keyword arguments. Here is the code to clarify:
change_hash({rand1: 'om'})
# ArgumentError: unknown keyword: rand1
To work around this you can pass two separate hashes into the method with second one (the one for keyword arguments) being empty:
def change_hash(h={}, rand: 'om')
h['hey'] = true
end
k = {}
change_hash(k, {})
k
#=> {'hey' => true}
From the practical point of view it is better to avoid metdhod signature like that in production code, because it is very easy to make an error while using the method.
Related
the following code return this error:
block in find_word_lengths': undefined method `[]=' for 3:Integer (NoMethodError)
animals = ['cat', 'horse', 'rabbit', 'deer']
def find_word_lengths(word_list)
word_list.reduce(Hash.new()) do |result, animal|
result[animal] = animal.length
end
end
puts find_word_lengths(animals)
The return value of the block is the accumulator value for the next iteration. That is how a fold works.
Assignments in Ruby evaluate to the right-hand side. So, in the first iteration of reduce, the block evaluates to 3 (the length of 'cat'). Which means that in the second iteration of reduce, result is 3, and you are essentially running
3['horse'] = 5
# which is equivalent to
3.[]=('horse', 5)
Which is why you are getting the error message that the Integer 3 does not respond to the message []=.
So, you need to make sure that your block always returns the value that you want to use for the accumulator in the next iteration. Something like this:
word_list.reduce(Hash.new()) do |result, animal|
result.tap {|result| result[animal] = animal.length }
end
This would be the obvious solution, although somewhat cheating.
word_list.reduce(Hash.new()) do |result, animal|
result.merge(animal => animal.length)
end
Would be more idiomatic.
However, when you want to fold into a mutable object, it makes more sense to use Enumerable#each_with_object instead of Enumerable#reduce. each_with_object ignores the result of the block, and simply passes the same object every time. Note that somewhat confusingly, the order of the block parameters is swapped in each_with_object compared to reduce.
word_list.each_with_object(Hash.new()) do |animal, result|
result[animal] = animal.length
end
But I guess the most idiomatic solution would be something like this:
word_list.map {|word| [word, word.length] }.to_h
By the way, in Ruby, it is idiomatic to leave out the parentheses for the argument list if you are not passing any arguments, so Hash.new() should be Hash.new instead. Even more important than being idiomatic is to be consistent – confusingly, you leave out the parentheses for animal.length, but not for Hash.new
Even more idiomatically, you would use the Hash literal notation instead of the Hash::new method, i.e. you should use {} instead of Hash.new.
Consider the following:
(1..10).inject{|memo, n| memo + n}
Question:
How does n know that it is supposed to store all the values from 1..10? I'm confused how Ruby is able to understand that n can automatically be associated with (1..10) right away, and memo is just memo.
I know Ruby code blocks aren't the same as the C or Java code blocks--Ruby code blocks work a bit differently. I'm confused as to how variables that are in between the upright pipes '|' will automatically be assigned to parts of an object. For example:
hash1 = {"a" => 111, "b" => 222}
hash2 = {"b" => 333, "c" => 444}
hash1.merge(hash2) {|key, old, new| old}
How do '|key, old, new|' automatically assign themselves in such a way such that when I type 'old' in the code block, it is automatically aware that 'old' refers to the older hash value? I never assigned 'old' to anything, just declared it. Can someone explain how this works?
The parameters for the block are determined by the method definition. The definition for reduce/inject is overloaded (docs) and defined in C, but if you wanted to define it, you could do it like so (note, this doesn't cover all the overloaded cases for the actual reduce definition):
module Enumerable
def my_reduce(memo=nil, &blk)
# if a starting memo is not given, it defaults to the first element
# in the list and that element is skipped for iteration
elements = memo ? self : self[1..-1]
memo ||= self[0]
elements.each { |element| memo = blk.call(memo, element) }
memo
end
end
This method definition determines what values to use for memo and element and calls the blk variable (a block passed to the method) with them in a specific order.
Note, however, that blocks are not like regular methods, because they don't check the number of arguments. For example: (note, this example shows the usage of yield which is another way to pass a block parameter)
def foo
yield 1
end
# The b and c variables here will be nil
foo { |a, b, c| [a,b,c].compact.sum } # => 1
You can also use deconstruction to define variables at the time you run the block, for example if you wanted to reduce over a hash you could do something like this:
# this just copies the hash
{a: 1}.reduce({}) { |memo, (key, val)| memo[key] = val; memo }
How this works is, calling reduce on a hash implicitly calls to_a, which converts it to a list of tuples (e.g. {a: 1}.to_a = [[:a, 1]]). reduce passes each tuple as the second argument to the block. In the place where the block is called, the tuple is deconstructed into separate key and value variables.
A code block is just a function with no name. Like any other function, it can be called multiple times with different arguments. If you have a method
def add(a, b)
a + b
end
How does add know that sometimes a is 5 and sometimes a is 7?
Enumerable#inject simply calls the function once for each element, passing the element as an argument.
It looks a bit like this:
module Enumerable
def inject(memo)
each do |el|
memo = yield memo, el
end
memo
end
end
And memo is just memo
what do you mean, "just memo"? memo and n take whatever values inject passes. And it is implemented to pass accumulator/memo as first argument and current collection element as second argument.
How do '|key, old, new|' automatically assign themselves
They don't "assign themselves". merge assigns them. Or rather, passes those values (key, old value, new value) in that order as block parameters.
If you instead write
hash1.merge(hash2) {|foo, bar, baz| bar}
It'll still work exactly as before. Parameter names mean nothing [here]. It's actual values that matter.
Just to simplify some of the other good answers here:
If you are struggling understanding blocks, an easy way to think of them is as a primitive and temporary method that you are creating and executing in place, and the values between the pipe characters |memo| is simply the argument signature.
There is no special special concept behind the arguments, they are simply there for the method you are invoking to pass a variable to, like calling any other method with an argument. Similar to a method, the arguments are "local" variables within the scope of the block (there are some nuances to this depending on the syntax you use to call the block, but I digress, that is another matter).
The method you pass the block to simply invokes this "temporary method" and passes the arguments to it that it is designed to do. Just like calling a method normally, with some slight differences, such as there are no "required" arguments. If you do not define any arguments to receive, it will happily just not pass them instead of raising an ArgumentError. Likewise, if you define too many arguments for the block to receive, they will simply be nil within the block, no errors for not being defined.
Given this method definition:
def foo(a = nil, b: nil)
p a: a, b: b
end
When I invoke the method with a single hash argument, the hash is always implicitly converted to keyword arguments, regardless of **:
hash = {b: 1}
foo(hash) #=> {:a=>nil, :b=>1}
foo(**hash) #=> {:a=>nil, :b=>1}
I can pass another (empty) hash as a workaround:
foo(hash, {}) #=> {:a=>{:b=>1}, :b=>nil}
But, this looks pretty cumbersome and awkward.
I would have expected Ruby to handle this more like arrays are handled, i.e.:
foo(hash) #=> {:a=>{:b=>1}, :b=>nil}
foo(**hash) #=> {:a=>nil, :b=>1}
And using literals:
foo({b: 1}) #=> {:a=>{:b=>1}, :b=>nil}
foo(b: 1) #=> {:a=>nil, :b=>1}
foo(**{b: 1}) #=> {:a=>nil, :b=>1}
The current implementation looks like a flaw and the way I was expecting it to work seems obvious.
Is this an overlooked edge case? I don't think so. There's probably a good reason that it wasn't implemented this way.
Can someone enlighten me, please?
As for the lack of ** part:
My guess is that, to make method invocation simple, Ruby always once interprets the key: value form without the braces as a hash with omitted braces, whether it is actually going to be interpreted as such hash or as keyword arguments.
Then, in order to interpret that as keyword arguments, ** is implicitly applied to it.
Therefore, if you had passed an explicit hash, it will not make difference to the process above, and there is room for it to be interpreted either as an actual hash or as keyword arguments.
What happens when you do pass ** explicitly like:
method(**{key: value})
is that the hash is decomposed:
method(key: value)
then is interpreted as a hash with omitted braces:
method({key: value})
then is interpreted either as a hash or as a keyword argument.
As for keyword arguments having priority over other arguments, see this post on Ruby core: https://bugs.ruby-lang.org/issues/11967.
I have the following method signature
def invalidate_cache(suffix = '', *args)
# blah
end
I don't know if this is possible but I want to call invalidate_cache and omit the first argument sometimes, for example:
middleware.invalidate_cache("test:1", "test")
This will of course bind the first argument to suffix and the second argument to args.
I would like both arguments to be bound to args without calling like this:
middleware.invalidate_cache("", "test:1", "test")
Is there a way round this?
Use keyword arguments (this works in Ruby 2.0+):
def invalidate_cache(suffix: '', **args) # note the double asterisk
[suffix, args]
end
> invalidate_cache(foo: "any", bar: 4242)
=> ["", {:foo=>"any", :bar=>4242}]
> invalidate_cache(foo: "any", bar: 4242, suffix: "aaaaa")
=> ["aaaaa", {:foo=>"any", :bar=>4242}]
Note that you will have the varargs in a Hash instead of an Array and keys are limited to valid Symbols.
If you need to reference the arguments by position, create an Array from the Hash with Hash#values.
How about you create a wrapper method for invalidate_cache that just calls invalidate_cache with the standard argument for suffix.
In order to do this, your code has to have some way of telling the difference between a suffix, and just another occurrence of args. E.g. In your first example, how is your program supposed to know that you didn't mean for "test:1" to actually be the suffix?
If you can answer that question, you can write some code to make the method determine at run time whether or not you provided a suffix. For example, say you specify that all suffixes have to start with a period (and no other arguments will). Then you could do something like this:
def invalidate_cache(*args)
suffix = (args.first =~ /^\./) ? args.shift : ''
[suffix, args]
end
invalidate_cache("test:1", "test") #=> ["", ["test:1", "test"]]
invalidate_cache(".jpeg", "test:1", "test") #=> [".jpeg", ["test:1", "test"]]
If, however, there actually is no way of telling the difference between an argument meant as a suffix and one meant to be lumped in with args, then you're kind of stuck. You'll either have to keep passing suffix explicitly, change the method signature to use keyword arguments (as detailed in karatedog's answer), or take an options hash.
I'm trying to return a list of values based on user defined arguments, from hashes defined in the local environment.
def my_method *args
#initialize accumulator
accumulator = Hash.new(0)
#define hashes in local environment
foo=Hash["key1"=>["var1","var2"],"key2"=>["var3","var4","var5"]]
bar=Hash["key3"=>["var6"],"key4"=>["var7","var8","var9"],"key5"=>["var10","var11","var12"]]
baz=Hash["key6"=>["var13","var14","var15","var16"]]
#iterate over args and build accumulator
args.each do |x|
if foo.has_key?(x)
accumulator=foo.assoc(x)
elsif bar.has_key?(x)
accumulator=bar.assoc(x)
elsif baz.has_key?(x)
accumulator=baz.assoc(x)
else
puts "invalid input"
end
end
#convert accumulator to list, and return value
return accumulator = accumulator.to_a {|k,v| [k].product(v).flatten}
end
The user is to call the method with arguments that are keywords, and the function to return a list of values associated with each keyword received.
For instance
> my_method(key5,key6,key1)
=> ["var10","var11","var12","var13","var14","var15","var16","var1","var2"]
The output can be in any order. I received the following error when I tried to run the code:
undefined method `assoc' for #<Hash:0x10f591518> (NoMethodError)
Please would you point me how to troubleshoot this? In Terminal assoc performs exactly how I expect it to:
> foo.assoc("key1")
=> ["var1","var2"]
I'm guessing you're coming to Ruby from some other language, as there is a lot of unnecessary cruft in this method. Furthermore, it won't return what you expect for a variety of reasons.
`accumulator = Hash.new(0)`
This is unnecessary, as (1), you're expecting an array to be returned, and (2), you don't need to pre-initialize variables in ruby.
The Hash[...] syntax is unconventional in this context, and is typically used to convert some other enumerable (usually an array) into a hash, as in Hash[1,2,3,4] #=> { 1 => 2, 3 => 4}. When you're defining a hash, you can just use the curly brackets { ... }.
For every iteration of args, you're assigning accumulator to the result of the hash lookup instead of accumulating values (which, based on your example output, is what you need to do). Instead, you should be looking at various array concatenation methods like push, +=, <<, etc.
As it looks like you don't need the keys in the result, assoc is probably overkill. You would be better served with fetch or simple bracket lookup (hash[key]).
Finally, while you can call any method in Ruby with a block, as you've done with to_a, unless the method specifically yields a value to the block, Ruby will ignore it, so [k].product(v).flatten isn't actually doing anything.
I don't mean to be too critical - Ruby's syntax is extremely flexible but also relatively compact compared to other languages, which means it's easy to take it too far and end up with hard to understand and hard to maintain methods.
There is another side effect of how your method is constructed wherein the accumulator will only collect the values from the first hash that has a particular key, even if more than one hash has that key. Since I don't know if that's intentional or not, I'll preserve this functionality.
Here is a version of your method that returns what you expect:
def my_method(*args)
foo = { "key1"=>["var1","var2"],"key2"=>["var3","var4","var5"] }
bar = { "key3"=>["var6"],"key4"=>["var7","var8","var9"],"key5"=>["var10","var11","var12"] }
baz = { "key6"=>["var13","var14","var15","var16"] }
merged = [foo, bar, baz].reverse.inject({}, :merge)
args.inject([]) do |array, key|
array += Array(merged[key])
end
end
In general, I wouldn't define a method with built-in data, but I'm going to leave it in to be closer to your original method. Hash#merge combines two hashes and overwrites any duplicate keys in the original hash with those in the argument hash. The Array() call coerces an array even when the key is not present, so you don't need to explicitly handle that error.
I would encourage you to look up the inject method - it's quite versatile and is useful in many situations. inject uses its own accumulator variable (optionally defined as an argument) which is yielded to the block as the first block parameter.