红宝石块迭代

Question

我对ruby中的迭代感到困惑。在我写的以下代码中，我预计打印出来的两个路径应该是相同的。但实际上他们不是。似乎路径在for循环中已更改。

我的代码有问题吗？由于

def one_step_search(dest,paths) 
    direction = %w(north east south west) 
    new_paths = [] 
    paths.map do |path| 

    print "original path is: " 
    print_path path 

    curr_room = path.last 
    for i in 0..3 
     new_path = path 
     if !curr_room.send("exit_#{direction[i]}").nil? 
     next_room_tag = curr_room.send("exit_#{direction[i]}")[0] 
     next_room = find_room_by_tag(next_room_tag) 
     if !new_path.include?(next_room) # don't go back to the room visited before 
      new_path << next_room 
      new_paths << new_path 
      print "new path is: " 
      print_path path 
      return new_paths if dest.tag == next_room_tag 
     end 
     end 
    end 
    end 

    return new_paths 
end 

Answer 1

在我看来，这个问题是在这条线

new_path = path 

你可能会认为new_path和path是不同的对象，但事实并非如此。我将举例说明：

a = "foo" 
b = a 
puts a.sub!(/f/, '_') 
puts a     # => "_oo" 
puts b     # => "_oo" 

a和b是指向一个对象的引用。 你的simpliest解决方案将使用dup或clone

new_path = path.clone 

但实际上你的代码需要良好的清洁。