Ruby Error: private method `execute' called - ruby

Question

Edit: To save confusion, I've added the actual code I'm having the issue with instead of the example. Same error as before. This is an interactive shell, and for example if I type ls -a it attempts to call the 'execute' method, but instead crashes saying that the method is private. I don't understand how it's private, nor how to work around this.
$ ruby shell.rb
ls -a
Traceback (most recent call last):
3: from shell.rb:59:in `<main>'
2: from shell.rb:6:in `main'
1: from shell.rb:6:in `loop'
shell.rb:10:in `block in main': private method `execute' called for #<Command:0x00005637cb75b1e8 #args=["ls"#0, "-a"#3]> (NoMethodError)
Ruby Code:
require 'readline'
require 'parslet'
def main
loop do
cmdline = Readline.readline("> ", true)
# p tree
tree.execute
end
end
def parse_cmdline(cmdline)
raw_tree = Parser.new.parse(cmdline)
Transform.new.apply(raw_tree)
end
class Parser < Parslet::Parser
root :cmdline
rule(:cmdline) { command }
rule(:command) { arg.as(:arg).repeat(1).as(:command) }
rule(:arg) { match[%q{^\s}].repeat(1) >> space? }
rule(:space?) {space.maybe }
rule(:space) { match[%q{\s}].repeat(1).ignore }
end
class Transform < Parslet::Transform
rule(command: sequence(:args)) { Command.new(args) } #ls
rule(arg: simple(:arg)) { arg } # -a
end
class Command
def initialize(args)
#args = args
end
end
def execute
spawn(*#args)
end
main

Answer 1

Why it doesn't work?
tree = (3) # tree is an integer
tree.execute # integers don't have an execute method
It's unclear what you are trying to do?
Try this instead:
I think you are trying to create a tree class? It's hard to say.
class Tree
def initialize(tree)
#tree = tree
end
def execute
puts #tree
end
end
def test
loop do
tree = Tree.new(3)
tree.execute
end
end
test
Because you are doing an infinite loop, you'll have to press: "CTRL + C" to kill the program, when you want it to stop (That'll work for Ubuntu, but not sure how it will work with other linux based distros or macs or windows).