如何將塊傳遞給子方法？

Question

說我有以下代碼：

def a(n, m, &block)
  yield if block_given?
end

def a
  # My question is here. When a is called, block might be or might not be
  # given. Below line is obvious wrong. How to call b and properly pass 
  # block to b?
  b(1, 2, &block)
end

a  # call a without block

a { # call a with a block
    puts "in block"
}

Answer 1

編寫a()以接受一個塊。 它暗示是可選的，並且如安德魯·馬歇爾（Andrew Marshall）所指出的那樣，如果沒有給出，將以&nil形式傳遞。

def b(n, m, &block)
  yield if block_given?
  puts "no block" if !block_given?
end

def a( &block )
  b(1, 2, &block)
end

a  # call a without block

a { # call a with a block
    puts "in block"
}

輸出：

no block
in block