[ruby-core:115490] [Ruby master Bug#19877] Non intuitive behavior of syntax only applied to literal value

Issue #19877 has been updated by kddnewton (Kevin Newton).


We need an answer on this in order to properly understand what we need to support. In particular,

``` ruby
(1; /(?<foo>)/) =~ s
```

assigning to the local seems incorrect. It would mean the only way for a parser (prism or any other) to be correct would be to mirror CRuby dropping literal nodes from the tree. In the meantime I know @yui-knk has been working on adding those nodes back in for the universal parser.

So either way, we need to know what is "correct" in this case.


----------------------------------------
Bug #19877: Non intuitive behavior of syntax only applied to literal value
https://bugs.ruby-lang.org/issues/19877#change-105420

* Author: tompng (tomoya ishida)
* Status: Closed
* Priority: Normal
* ruby -v: ruby 3.3.0dev (2023-09-08T23:08:32Z master b635a66e95) [x86_64-linux]
* Backport: 3.0: UNKNOWN, 3.1: UNKNOWN, 3.2: UNKNOWN
----------------------------------------
Non intuitive behavior of syntax only applied to literal value

Some ruby syntax is only applied to literal value.
~~~ruby
def 1.foo; end # receiver is a literal, it is Syntax Error
/(?<a>)/ =~ s # receiver is regexp literal, it will assign to local variable
if cond1..cond2; end # range-like syntax appears in condition, it is flipflop
~~~

If it is wrapped with parenthesis, the behavior seems not intuitive for me, and YARP parses it differently.
~~~ruby
def (1).foo; end # Syntax Error
def ((1;1)).foo; end # Syntax Error
def ((;1)).foo; end # Syntax OK
def ((1+1;1)).foo; end # Syntax OK
def ((%s();1)).foo; end # Syntax Error
def ((%w();1)).foo; end # Syntax OK
def ("#{42}").foo; end # Syntax Error
def (:"#{42}").foo; end # Syntax OK

(/(?<a>)/) =~ s # assigns to a
(;/(?<a>)/) =~ s # does not assigns
(%s();/(?<a>)/) =~ s # assigns to a
(%w();/(?<a>)/) =~ s # does not assigns
(1; (2; 3; (4; /(?<a>)/))) =~ s # assigns to a
(1+1; /(?<a>)/) =~ s # does not assign

if ((cond1..cond2)); end # flipflop
if (; cond1..cond2); end # range
if (1; cond1..cond2); end # flipflop
if (%s(); cond1..cond2); end # flipflop
if (%w(); cond1..cond2); end # range
if (1; (2; (3; 4; cond1..cond2))); end # flipflop
if (1+1; cond1..cond2); end # range
~~~

I expect YARP and parse.y parses same.
I expect all parenthesis-wrapped result same.
I think it is simple and intuitive if parenthesis-wrapped code always behaves different from non-wrapped code because there are more complex variation like this
~~~ruby
def (class A; 1; end).foo; end
(break; /?<a>/) =~ s
class A; /?<a>/; end =~ s
~~~



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