ruby - 如何将 block 传递给子方法？

Question

假设我有以下代码:

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

编写 a() 来接受一个 block 。它暗示是可选的，正如安德鲁编码(marshal)指出的那样，如果没有给出，将作为 &nil 传递。

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"
}

输出:

no block
in block