Bug #15880: Wrong precedence of the if modifier in pattern matching

• Status: Rejected
• Assignee: -
• Target version: -
• ruby -v: ruby 2.7.0dev (2019-05-20 trunk ab0f2deab1) [x86_64-darwin18]
• Backport: 2.4: UNKNOWN, 2.5: UNKNOWN, 2.6: UNKNOWN

[ibylich (Ilya Bylich)]

When "If" is used as an "If modifier" it runs before the expression that it wraps:

=> puts 1 if (puts 2; true)
2
1

However, when it's used in the pattern matching destructuring runs first:

class A
  def deconstruct
    puts 'deconstruct called'
    [1]
  end
end

p case A.new
in A[1] if (puts 'if check'; true)
  'yes'
else
  'no'
end

# prints
# deconstruct called
# if check
# "yes"

I personally think that it's very confusing as it doesn't reflect the code. I assumed it to not run destructuring if the check returns false (especially because destructuring is allowed to have side-effects)

[marcandre (Marc-Andre Lafortune)]

• Status changed from Open to Rejected

This is necessary so the if can depend on the variables of the matching, e.g. in [Integer => x] if x.odd?

[duerst (Martin Dürst)]

It's clear that the order of evaluation has to be the way it is currently. But in this case, if just sounds wrong, not only because of examples such as puts 1 if condition, but also because of general English.

So I think changing if to something else is highly desirable. But I unfortunately don't have any good proposal.