如何在ruby中混合必需的参数和可选参数？
我正在研究ruby，以下是我的代码片段。 我尝试混合必需的参数和选项参数。 但它给了我一个错误：
def my_method(a, b, *p, d) puts a puts b puts "content of p is :" + p.to_s puts d end my_method(10, 20, 34, 45) 错误代码段：
syntax error def my_method(a, b, *p, d) ^
这个错误的原因是什么？
注意：我使用“ruby 1.8.5”ruby版本
您在任何最新版本的Ruby中都可以使用这些代码。 但是，它不适用于真正旧版本的Ruby，因为splat参数之后需要的位置参数仅在Ruby 1.9.0中引入。
Ruby中参数列表的伪正则表达式是这样的：
mand* opt* splat? mand* (mand_kw | opt_kw)* kwsplat? block?
这是一个例子：
def foo(m1, m2, o1=:o1, o2=:o2, *splat, m3, m4, ok1: :ok1, mk1:, mk2:, ok2: :ok2, **ksplat, &blk) Hash[local_variables.map {|var| [var, eval(var.to_s)] }] 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]] 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: # }
数组splats（ *p ）仅作为参数列表中的最后一个参数有效，因为它意味着“将传递给该方法的所有其余参数收集到数组p ”。
只需从方法签名中删除d ，在您的方法中，只需设置d = p.pop即可。 或者，在*p之前移动d 。