了解ruby方法$ stdout的工作方式

Question

Here is a simple ruby script, that takes input from a user and provides an output(yes, it will be refactored). I wanted this script to provide output into a text file, rather than console window. That was accomplished, by simply adding $stdout = File.new('out.txt', 'w') , but I thought that this line will just describe a variable which I will use later to tell script to use it to write output into created file.

I cant find much documentation about this method and wondering how does this program knows how to write generated output into that file?

Answer 1

$stdout is a global variable. By default it stores an object of type IO associated with the standard output of the program (which is, by default, the console).

puts is a method of the Kernel module that actually calls $stdout.send() and pass it the list of arguments it receives. As the documentation explains, puts(obj, ...) is equivalent to $stdout.puts(obj, ...) .

Your code replaces $stdout with an object of type File that extends class IO . When it is created, your object opens the file out.txt for writing and together with its inheritance from IO it is fully compatible with the default behaviour of $stdout .

Since by default, all the output goes to $stdout , your new definition of $stdout ensures the output is written to the file out.txt without other changes in the code.

Answer 2

$stdout is a global variable (as indicated by $ ) and according to the documentation, puts is:

Equivalent to

 $stdout.puts(obj, ...) 

If you assign another object to $stdout , then Kernel#puts will simply send puts to that object. Likewise, print will send write :

class Foo < BasicObject
  def puts(*args)
    ::STDOUT.puts "Foo#puts called with #{args.inspect}"
  end

  def write(*args)
    ::STDOUT.puts "Foo#write called with #{args.inspect}"
  end
end

$stdout = Foo.new

puts 'hello', 'world'
# Foo#puts called with ["hello", "world"]
print "\n"
# Foo#write called with ["\n"]

Note that if you assign to $stdout , Ruby checks whether the object responds to write . If not, a TypeError will be raised.

Answer 3

It's probably worth highlighting the fact, mentioned by the other posts, that this is a global variable, and modifying it is not thread safe. For instance, if you point $stdout to a new IO stream, anything across the app (on that process thread) that would be logged to stdout will now be logged to your new IO stream. This can lead to multiple streams of unexpected and possibly sensitive input landing in your new IO stream.