Feature #11737: Pass in expression to then block in `case expression`

• Status: Rejected
• Assignee: -
• Target version: -

[danielpclark (Daniel P. Clark)]

Ruby's case <expression> scenario allows the expression to be tested in the when test with :===, but in the then block the expression is not accessible. Only if the expression has been assigned to a variable beforehand can it be checked.

case 4
when ->i{ puts :when; true}
  ->i{ puts i}
else
  :foo
end
# when
#  => #<Proc:0x00000000d91e58@(irb):16 (lambda)> 

case 4
when ->i{ puts :when; true}
  puts _
else
  :foo
end
# when
# #<Proc:0x00000000d91e58@(irb):16 (lambda)>
#  => nil 

case 4
when 4
then _
end
#  => nil

case 4
when 4
then ->i{puts i}
end
#  => #<Proc:0x000000015f9be0@(irb):36 (lambda)> 

If some one wanted to give an expression after case that wasn't assigned to a variable then there is no access to it in the then block.

I suggest assigning the expression to the _ variable during a case/when/then scenario. Here's a rather contrived example use case.

case Enumerator.new do |y| y << 1; y << 2; y << 3; end
when ->e{ 2.times e.next; true}
then _.peek
end == 3

[danielpclark (Daniel P. Clark)]

Please fix the when scenario in the last example

case Enumerator.new do |y| y << 1; y << 2; y << 3; end
when ->e{ 2.times{e.next}; true}
then _.peek
end == 3

[danielpclark (Daniel P. Clark)]

Since https://bugs.ruby-lang.org/issues/11734#note-3 has informed me the _ is an IRB feature and Minitest uses _() in its test suite perhaps we should not use _ for the case/when/then.

How about __case__

case Enumerator.new do |y| y << 1; y << 2; y << 3; end
when ->e{ 2.times{e.next}; true}
then __case__.peek
end == 3

[nobu (Nobuyoshi Nakada)]

It reminded me a rejected proposal:

case expr
when matcher => result
  ...
end

[danielpclark (Daniel P. Clark)]

Nobuyoshi Nakada wrote:

It reminded me a rejected proposal:

case expr
when matcher => result
  ...
end

I don't think so. It's not too different from what already works if you create a variable beforehand.

e = Enumerator.new do |y| y << 1; y << 2; y << 3; end

case e
when ->x{ 2.times{x.next}; true}
then e.peek
end == 3
# => true

I just found it odd that in the when clause I can use a proc to grab the current case item, but in the then block there is nothing to check unless a variable has already been assigned.

[wanabe (_ wanabe)]

Nobuyoshi Nakada wrote:

It reminded me a rejected proposal:

case expr
when matcher => result
  ...
end

Is the proposal [ ruby-dev:17615 ] ?
Or other?

[danielpclark (Daniel P. Clark)]

case 4 # __case__ = 4

when 4 # 4 === __case__

puts "Success! #{__case__} is what we wanted!" # __case__ not defined! ... where did it go?

end

[matz (Yukihiro Matsumoto)]

• Status changed from Open to Rejected

It's not worth adding special expression or variable for retrieving the when value.
Just use assignment.

Matz.

[danielpclark (Daniel P. Clark)]

Thank you. I appreciate your consideration.