I want to be able to have one ruby file that can require all the common dependencies, so that other files can just have one require on this shared file.for example; I have foo.rb and bar.rb and allrequires.rb. I want to have the line require "allrequires.rb" in both foo.rb and bar.rb, but bar.rb doesn't need all the requires.
Does it matter if I use require in .rb file that do not really require that file? Could it have an impact on performance maybe?
I am currently on ruby 1.8.7 (2010-08-16 patchlevel 302) [i386-mingw32]
It looks like it is not the best idea to 'share'/use all requires in both .rb files. What would be solution to that?
Right now I can think of using file name in a condition.
There's two main performance penalties:
require itself. In Ruby 1.9.1 and Ruby 1.9.2, the time taken to do all the requires had worse than linear scalability - if you doubled the number of requires, it took you more than twice as long - I think it took you four times as long.required. If you have code like the following, then executing the code will take a non-trivial amount of time.class MyClass
MY_CONSTANT = File.read("data.txt")
end
Another approach of conditional requiring, the following script gives no error on the JSON parser because it is named require1.rb, in scripts that have no name like require1.rb of script2.rb the gem isn't required
require 'json' if "require1.rb, script2.rb"[File.basename(__FILE__)]
p File.basename(__FILE__)
text = '[{ "name" : "car", "status": "good"}, { "name" : "bus", "status": "bad"},{ "name" : "taxi", "status": "soso"},
{"noname":"", "status" : "worse"}
]'
data = JSON.parse(text)
p data.collect { |item| item['name'] }
EDIT: here a version that uses an array
["require1.rb","script1.rb"].find{|script|require 'json' if script===File.basename(__FILE__)}
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