在Ruby中,如何从类的实例中调用类方法?说我有
class Truck
def self.default_make
# Class method.
"mac"
end
def initialize
# Instance method.
Truck.default_make # gets the default via the class's method.
# But: I wish to avoid mentioning Truck. Seems I'm repeating myself.
end
end
Truck.default_make行检索默认值。但有办法不提卡车吗?似乎应该有。
你做得对。类方法(类似于c++或Java中的“静态”方法)不是实例的一部分,因此必须直接引用它们。
关于这一点,在你的例子中,你最好将'default_make'设置为常规方法:
#!/usr/bin/ruby
class Truck
def default_make
# Class method.
"mac"
end
def initialize
# Instance method.
puts default_make # gets the default via the class's method.
end
end
myTruck = Truck.new()
类方法对于使用该类的实用程序类型函数更有用。例如:
#!/usr/bin/ruby
class Truck
attr_accessor :make
def default_make
# Class method.
"mac"
end
def self.buildTrucks(make, count)
truckArray = []
(1..count).each do
truckArray << Truck.new(make)
end
return truckArray
end
def initialize(make = nil)
if( make == nil )
@make = default_make()
else
@make = make
end
end
end
myTrucks = Truck.buildTrucks("Yotota", 4)
myTrucks.each do |truck|
puts truck.make
end
如果你可以访问委托方法,你可以这样做:
[20] pry(main)> class Foo
[20] pry(main)* def self.bar
[20] pry(main)* "foo bar"
[20] pry(main)* end
[20] pry(main)* delegate :bar, to: 'self.class'
[20] pry(main)* end
=> [:bar]
[21] pry(main)> Foo.new.bar
=> "foo bar"
[22] pry(main)> Foo.bar
=> "foo bar"
或者,如果你有一个或两个以上的方法,你想委托给类和实例:
[1] pry(main)> class Foo
[1] pry(main)* module AvailableToClassAndInstance
[1] pry(main)* def bar
[1] pry(main)* "foo bar"
[1] pry(main)* end
[1] pry(main)* end
[1] pry(main)* include AvailableToClassAndInstance
[1] pry(main)* extend AvailableToClassAndInstance
[1] pry(main)* end
=> Foo
[2] pry(main)> Foo.new.bar
=> "foo bar"
[3] pry(main)> Foo.bar
=> "foo bar"
提醒一句:
不要只是随机地将所有不改变状态的东西委托给类和实例,因为这会开始遇到奇怪的名称冲突问题。这样做是有节制的,只有在你检查没有其他压扁。