在Elixir中,我如何记录函数将返回实现特定行为的模块?
要使用一个简单的例子,假设我创建了一个GreeterBehaviour由两个模块实现的行为:
defmodule GreeterBehaviour do
@callback say_hello(String.t) :: String.t
end
defmodule FormalGreeter do
@behaviour GreeterBehaviour
def say_hello(name) do
"Good day to you #{name}"
end
end
defmodule CasualGreeter do
@behaviour GreeterBehaviour
def say_hello(name) do
"Hey #{name}"
end
end
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然后,我想通过函数检索Greeter,轻松地换掉这些实现中的任何一个:
defmodule MyApp do
def main do
greeter().say_hello("Pete") |> IO.puts
end
@spec greeter() :: GreeterBehaviour # This doesn't work with dialyzer
def greeter do
FormalGreeter # Can easily be swapped to CasualGreeter
end
end
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Dialyzer将成功检查两者CasualGreeter并FormalGreeter正确实现该GreeterBehaviour行为.但是,我如何定义一个typespec,以便Dialyzer检查greeter/0返回一个实际上实现的模块GreeterBehaviour?
使用@spec greeter() :: GreeterBehaviour不起作用,因为Dialyzer会发出警告:
lib/my_app.ex:19: Invalid type specification for function 'Elixir.MyApp':greeter/0. The success typing is () -> 'Elixir.FormalGreeter'
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小智 1
在你的行为中,你可以定义一个类型say_hello:
@type f :: (String.t() -> String.t())
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你的greeter函数可以返回一个模块+函数:&module.say_hello/1
规格是:
@spec greeter() :: GreeterBehaviour.f()
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defmodule GreeterBehaviour do
@type f :: (String.t() -> String.t())
@callback say_hello(String.t()) :: String.t()
end
defmodule FormalGreeter do
@behaviour GreeterBehaviour
def say_hello(name) do
"Good day to you #{name}"
end
end
defmodule CasualGreeter do
@behaviour GreeterBehaviour
def say_hello(name) do
"Hey #{name}"
end
end
defmodule MyApp do
def main do
greeter().("Pete") |> IO.puts()
end
# This will work with dialyzer
@spec greeter() :: GreeterBehaviour.f()
def greeter do
# Can easily be swapped to CasualGreeter
&FormalGreeter.say_hello/1
end
end
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