I have a method that looks like this:
class A
def my_method(a: 'key value', b = 'opt value')
# ...
end
end
Using reflection, I can get the parameter names like this:
A.new.method(:my_method).parameters
# => [[:key, :a], [:opt, :b]]
How can I get the default values of these parameters without invoking my_method?
This is not possible. I can't find the thread right now, but matz has explicitly said that this is by design. The problem is that the default value is an arbitrary expression, and since everything in Ruby is an expression, it can be anything.
For example, what about this:
def foo(bar = if rand < 0.5 then Time.now else rand end)
What would your proposed method return here?
Basically, you have two choices:
Evaluate the default value expression. This means you will get some value, but that doesn't tell you much about what the real expression is.
Don't evaluate the default value expression, by packaging it up in a Proc. But there's no way to get the expression out of a Proc again.
So, either way, you don't actually get any useful information.
Related
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.
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.
I will try to explain the problem with a simple example:
def enclose(x)
[x]
end
In my application, enclose does something more complex, but in essence it returns an array, the content of which is solely determined by the value of the parameter x. I could it use it like this:
foo = 'abcd'
....
foo = enclose(foo)
Now to my question: Is it possible to write a method enclose!, which simply replaces the parameter by its enclosed version, so that the example could be written as
foo = 'abcd'
....
enclose!(foo)
Since Ruby passes arguments by reference, I thought hat this could maybe be possible. The naive approach,
def enclose!(x)
x = [x]
end
does not work - I think this is because the assignment creates a new object and leaves the actual parameter untouched.
Is there way, that I can achieve my goal? I think in Smallalk, there would be a method become which would change the object identity, but I didn't find something similar in Ruby.
Since Ruby passes arguments by reference, I thought hat this could maybe be possible.
Ruby is pass-by-value, not pass-by-reference, which you have proven yourself, because otherwise your code would have worked.
I think in Smallalk, there would be a method become which would change the object identity, but I didn't find something similar in Ruby.
There isn't. Ruby has neither pass-by-reference nor become:, what you want simply isn't possible.
There's some other interesting posts about how ruby is pass by value, but the values are references.
What it boils down to is, you can modify the variable an object refers to, but you cannot change it to refer to another object.
> a = [1]
=> [1]
> def add_a(array)
> array << "a"
> end
=> :add_a
> add_a a
=> [1, "a"]
> a
=> [1, "a"]
There is a way to sort of accomplish what you are asking for but it's not quite pretty. Ruby has this concept of a binding (http://ruby-doc.org/core-2.2.0/Binding.html), which is like a CallContext in .NET.
You can do something like this:
def enclose(x)
[x]
end
def enclose!(x, binding)
eval("#{x} = [#{x}]", binding)
end
foo = 'abcd'
enclose!(:foo, binding)
=> ["abcd"]
In the script above, the :foo means you are passing the name of the variable, and the binding (context) where to find its value. Then you're dynamically calling eval to evaluate the assignment operation foo = [foo].
I know a bit about ruby way to handle objects and references. The replace stuff, ect ...
I know it d'ont work on fixnum, cause the var is the fixnum. But i wish to change the value of a fixnum inside a function, and that the value changed in the ouside var.
How can i do this ?
I guess i can use a string like this "1" but that's quite dirty.
Ruby will always pass-by-reference (because everything is an object) but Fixnum lacks any methods that allow you to mutate the value. See "void foo(int &x) -> Ruby? Passing integers by reference?" for more details.
You can either return a value that you then assign to your variable, like so:
a = 5
def do_something(value)
return 1 #this could be more complicated and depend on the value passed in
end
a = do_something(a)
or you could wrap your value in an object such as a Hash and have it updated that way.
a = {:value => 5}
def do_something(dict)
dict[:value] = 1
end
do_something(a) #now a[:value] is 1 outside the function
Hope this helps.
You could pass an array with a single number, like [1], or a hash like {value: 1}. Less ugly than a string, as your number itself remains a number, but less overhead than a new class...
When I was building a game I had the same problem you have. There was a numeric score that represented how many zombies you've killed and I needed to manually keep it in sync between Player (that incremented the score), ScoreBar and ScoreScreen (that displayed the score). The solution I've found was creating a separate class for the score that will wrap the value and mutate it:
class Score
def initialize(value = 0)
#value = value
end
def increment
#value += 1
end
def to_i
#value
end
def to_s
#value.to_s
end
end
Trying to do something weird that might turn into something more useful, I tried to define my own []= operator on a custom class, which you can do, and have it return something different than the value argument, which apparently you can't do. []= operator's return value is always value; even when you override this operator, you don't get to control the return value.
class Weird
def []=(key, value)
puts "#{key}:#{value}"
return 42
end
end
x = Weird.new
x[:a] = "a"
output "a:a"
return value => "a" # why not 42?
Does anyone have an explanation for this? Any way around it?
ruby MRI 1.8.7. Is this the same in all rubys; Is it part of the language?
Note that this behavior also applies to all assignment expressions (i.e. also attribute assignment methods: def a=(value); 42; end).
My guess is that it is designed this way to make it easy to accurately understand assignment expressions used as parts of other expressions.
For example, it is reasonable to expect x = y.a = z[4] = 2 to:
call z.[]=(4,2), then
call y.a=(2), then
assign 2 to the local variable x, then finally
yield the value 2 to any “surrounding” (or lower precedence) expression.
This follows the principle of least surprise; it would be rather surprising if, instead, it ended up being equivalent to x = y.a=(z.[]=(4,2)) (with the final value being influenced by both method calls).
While not exactly authoritative, here is what Programming Ruby has to say:
Programming Ruby (1.8), in the Expressions section:
An assignment statement sets the variable or attribute on its left side (the lvalue) to refer to the value on the right (the rvalue). It then returns that value as the result of the assignment expression.
Programming Ruby 1.9 (3rd ed) in section 22.6 Expressions, Conditionals, and Loops:
(right after describing []= method calls)
The value of an assignment expression is its rvalue. This is true even if the assignment is to an attribute method that returns something different.
It’s an assignment statement, and those always evaluate to the assigned value. Making this different would be weird.
I suppose you could use x.[]= :a, "a" to capture the return value.