Bug #16983: RubyVM::AbstractSyntaxTree.of(method) returns meaningless node if the method is defined in eval

• Status: Open
• Priority: Normal
• Assignee: -
• Target version: -
• ruby -v: ruby 2.8.0dev (2020-06-23T13:58:26Z master dc351ff984) [x86_64-linux]
• Backport: 2.5: UNKNOWN, 2.6: UNKNOWN, 2.7: UNKNOWN

[pocke (Masataka Kuwabara)]

Problem

RubyVM::AST.of(method) returns a meaningless node if the method is defined in eval.
For example:

p 'blah'

eval <<~RUBY, binding, __FILE__, __LINE__ + 1
  def foo
  end
RUBY

method = method(:foo)
pp RubyVM::AbstractSyntaxTree.of(method)
# => (STR@3:5-3:12 "def foo\n" + "end\n")

I expect the node of foo method, or nil. But it returns a STR node.

It becomes a big problem when AST.of receives arbitrary methods.
Because we can't distinguish a method is defined in eval or not.
It means we can't believe the returned value of AST.of if the method may receive a method defined in eval.

For example:

def do_something_for_each_method_ast(klass)
  klass.instance_methods(false).each do |m|
    ast = RubyVM::AbstractSyntaxTree.of(klass.instance_method(m))
    next unless ast

    do_something ast
  end
end

class A
  eval <<~RUBY, binding, __FILE__, __LINE__ + 1
    def foo
    end
  RUBY
end

do_something_for_each_method_ast A

In the example, I expect the do_something method receives only node for a method definition,
but it may pass a wrong node if any method is defined in eval.

Cause (I guess)

I guess the cause is misleading node number.
In and out of an eval block uses different sequences of node number.
So if I specify __FILE__ to eval, the actual file and code in eval may have the same node number.

For example

p 'blah' # Node number for 'blah' is 1, file name is "test.rb"

eval <<~RUBY, binding, __FILE__, __LINE__ + 1
  def foo # Node number for `def` is also 1, file name is also "test.rb"
  end
RUBY

method = method(:foo)
# It finds a node from node number 1 by reading "test.rb", so it get the str node.
pp RubyVM::AbstractSyntaxTree.of(method)
# => (STR@3:5-3:12 "def foo\n" + "end\n")