Question

假设我有一个带有块的函数with_foo，并将其包裹在一段代码中，例如

with_foo do
  puts "hello!"
end

现在我想将包装条件化，如

if do_with_foo?
  with_foo do
    puts "hello!"
  end
else
  puts "hello!" # without foo
end

有没有办法写这个更短/更优雅，这意味着无需重复代码puts "hello!"？

Answer 1

如果您愿意用块指定参数，则可以。

上面给出with foo，您可以编写这样的代码段：

whatever = proc {puts "hello"}
#build a proc object with a block
if do_with_foo?
  with_foo &whatever
#pass it to with_foo
else
  whatever.call
#normally call it
end

Answer 2

使用代理模式的概念证明：

class BlockWrapper
  def initialize(obj, use_wrapper)
    @obj = obj
    @use_wrapper = use_wrapper
  end

  def method_missing(*args, &block)
    @use_wrapper ? @obj.send(*args, &block) : block.call
  end
end

module Kernel
  def wrap_if(use_wrapper)
    BlockWrapper.new(self, use_wrapper)        
  end
end

def with_foo
  puts "with_foo: start"
  yield
  puts "with_foo: end"
end

wrap_if(true).with_foo do 
  puts "hello!"
end

wrap_if(false).with_foo do 
  puts "hello, no with_foo here!"
end

输出：

with_foo: start
hello!
with_foo: end
hello, no with_foo here!

Answer 3

我认为你可以这样做：

def without_foo &pr
  pr.call
end

send(do_with_foo?? "with_foo" : "without_foo") do
  puts "hello!"
end

Answer 4

您可以将复制代码放入Proc对象中，并将其作为块传递给方法或直接调用。

hello = proc { puts 'hello' }

with_foo(&hello)
# OR
hello.call

Answer 5

def simple_yielder
  yield
end
def maybe_with condition, with_method
  send( condition ? with_method : :simple_yielder ) { yield }
end

#...

maybe_with( do_with_foo?, :with_foo ) do
  puts "Hello?"
end

Ruby中的条件块

5 个答案: