隐式地将变量传递给Ruby中的块
我想重写这个片段:隐式地将变量传递给Ruby中的块,ruby,Ruby,我想重写这个片段: Chips.fix_game(324565) do |game_id| player1.chips.gain(game_id, 200) # chips qty player2.chips.lose(game_id, 200) end 这样做: Chips.fix_game(324565) do player1.chips.gain(200) player2.chips.lose(200) end 以某种方式将game\u id传递给player1.chi
Chips.fix_game(324565) do |game_id|
player1.chips.gain(game_id, 200) # chips qty
player2.chips.lose(game_id, 200)
end
这样做:
Chips.fix_game(324565) do
player1.chips.gain(200)
player2.chips.lose(200)
end
以某种方式将game\u id
传递给player1.chips
API入口点
第二个片段更简洁,块内没有空间更改game\u id
如何将
game\u id
值从Chips.fix\u game
方法隐式传递到player1.Chips
对象?Chips
可以提供使用fix\u game
设置的当前游戏id
,例如:
class Chips
@@current_game_id = nil
def self.current_game_id
@@current_game_id
end
def self.fix_game(game_id)
previous_game_id = current_game_id
@@current_game_id = game_id
yield
ensure
@@current_game_id = previous_game_id
end
end
class Player
def gain(amount)
puts "gaining #{amount} chips in game #{Chips.current_game_id}"
end
end
player = Player.new
Chips.fix_game(1) do
player.gain(100)
end
#=> "gaining 100 chips in game 1"
Chips.fix_game(2) do
player.gain(200)
end
#=> "gaining 200 chips in game 2"
在yield
之后恢复上一个游戏id允许嵌套块:
Chips.current_game_id #=> nil
Chips.fix_game(1) do
Chips.current_game_id #=> 1
Chips.fix_game(2) do
Chips.current_game_id #=> 2
end
Chips.current_game_id #=> 1
end
Chips.current_game_id #=> nil
虽然这看起来很方便,但全局状态可能会导致难以调试的问题。保重。好的,我看见你了。我不需要筑巢芯片。不过是游戏。多线程呢?