Question

这在Ruby中是可能的：

class SomeClass
  def initialize(a, *b, c)
  end
end

但这不是：

class SomeClass
  def initialize(a, *b, c='anything here')
  end
end

为什么？

编辑：这个问题没有答案。在答案链接中，第一个答案是：

splat表示“用尽所有剩余的参数”但是你呢提供一个可选参数，那么解释器如何知道是否可以最后一个参数是“数字”splat或可选项的一部分 “选项”？

这不是一个有效的答案，因为根据这个逻辑，Ruby在splat之后也不会允许强制参数。

Answer 1

Ruby的参数绑定语义已经非常复杂了。考虑这种方法：

def foo(m1, m2, o1=:o1, o2=:o2, *splat, m3, m4, 
          ok1: :ok1, mk1:, mk2:, ok2: :ok2, **ksplat, &blk)
  local_variables.map {|var| [var, eval(var.to_s)] }.to_h
end

method(:foo).arity
# => -5

method(:foo).parameters
# => [[:req, :m1], [:req, :m2], [:opt, :o1], [:opt, :o2], [:rest, :splat], 
#     [:req, :m3], [:req, :m4], [:keyreq, :mk1], [:keyreq, :mk2], 
#     [:key, :ok1], [:key, :ok2], [:keyrest, :ksplat], [:block, :blk]]

乍一看，您能说出以下调用的结果吗？

foo(1, 2, 3, 4)

foo(1, 2, 3, mk1: 4, mk2: 5)

foo(1, 2, 3, 4, mk1: 5, mk2: 6)

foo(1, 2, 3, 4, 5, mk1: 6, mk2: 7)

foo(1, 2, 3, 4, 5, 6, mk1: 7, mk2: 8)

foo(1, 2, 3, 4, 5, 6, 7, mk1: 8, mk2: 9)

foo(1, 2, 3, 4, 5, 6, 7, 8, mk1: 9, mk2: 10)

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11)

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12)

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12, k3: 13)

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12, k3: 13, k4: 14)

foo(1, 2, 3, 4, 5, 6, 7, 8, 
      ok1: 9, ok2: 10, mk1: 11, mk2: 12, k3: 13, k4: 14) do 15 end

现在，假设在splat参数之后添加带有默认参数的可选参数到该列表。找到理智的语义并非不可能，但这可能会导致一些不明显的结果。

你能想出简单，理智，向后兼容和非令人惊讶的语义吗？

BTW：这里是顶部方法的备忘单：

foo(1, 2, 3, 4)
# ArgumentError: missing keywords: mk1, mk2

foo(1, 2, 3, mk1: 4, mk2: 5)
# ArgumentError: wrong number of arguments (3 for 4+)

foo(1, 2, 3, 4, mk1: 5, mk2: 6)
# => { m1: 1, m2: 2, o1: :o1, o2: :o2, splat: [], m3: 3, m4: 4, 
#      ok1: :ok1, mk1: 5, mk2: 6, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, mk1: 6, mk2: 7)
# => { m1: 1, m2: 2, o1: 3, o2: :o2, splat: [], m3: 4, m4: 5, 
#      ok1: :ok1, mk1: 6, mk2: 7, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, mk1: 7, mk2: 8)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [], m3: 5, m4: 6, 
#      ok1: :ok1, mk1: 7, mk2: 8, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, mk1: 8, mk2: 9)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5], m3: 6, m4: 7, 
#      ok1: :ok1, mk1: 8, mk2: 9, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, mk1: 9, mk2: 10)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: :ok1, mk1: 9, mk2: 10, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: 9, mk1: 10, mk2: 11, ok2: :ok2, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: 9, mk1: 10, mk2: 11, ok2: 12, ksplat: {}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12, k3: 13)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: 9, mk1: 10, mk2: 11, ok2: 12, ksplat: {k3: 13}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, ok1: 9, mk1: 10, mk2: 11, ok2: 12, k3: 13, k4: 14)
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: 9, mk1: 10, mk2: 11, ok2: 12, ksplat: {k3: 13, k4: 14}, 
#      blk: nil }

foo(1, 2, 3, 4, 5, 6, 7, 8, 
      ok1: 9, ok2: 10, mk1: 11, mk2: 12, k3: 13, k4: 14) do 15 end
# => { m1: 1, m2: 2, o1: 3, o2: 4, splat: [5, 6], m3: 7, m4: 8, 
#      ok1: 9, mk1: 10, mk2: 11, ok2: 12, ksplat: {k3: 13, k4: 14}, 
#      blk: #<Proc:0xdeadbeefc00l42@(irb):15> }

为什么我可以在Ruby中使用splat但不是可选的参数？

1 个答案: