在红宝石功能无效在红宝石功能无效(Invalid function in ruby)

2019-05-12 07:58发布

这是为什么功能无效?

def request(method='get',resource, meta={}, strip=true)

end

unexcpected ')' keyword_end期待

谢谢!

Answer 1:

在Ruby中,你不能与周围可选参数必需的参数。 运用

def request(resource, method='get', strip=true, meta={})
end

将解决这一问题。

作为一个思想实验,考虑到原有的功能

def request(method='get',resource, meta={}, strip=true)
end

如果我调用该方法作为request(object) ,期望的行为是相当明显的-呼叫与方法object作为resource的参数。 但是,如果我把它作为request('post', object) ? 红宝石需要理解的语义method来决定是否'post'methodresource ,以及是否objectresourcemeta 。 这超出了Ruby的语法分析器的范围,因此它只是抛出一个无效的功能错误。

一对夫妇的其他提示:

我也把最后的元参数,它允许您通过在没有大括号,如哈希选项:

request(object, 'get', true, foo: 'bar', bing: 'bang')

正如安迪·海登在评论中指出,下列函数工作:

def f(aa, a='get', b, c); end

这是比较好的做法是将在函数结束时,所有的可选参数,以避免遵循这样的函数调用所需的智力体操。



Answer 2:

你只能有一组可选参数的参数列表。

伪正则表达式在Ruby中的参数列表是这样的:

mand* opt* splat? mand* (mand_kw | opt_kw)* kwsplat? block?

下面是一个例子:

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: #<Proc:0xdeadbeefc00l42@(irb):15> }

[注:强制关键字参数将在Ruby的2.1推出,其余全部已经工作。]



Answer 3:

尝试重新安排你的参数:

def Request(resource,strip=true,method='get',meta={})
end


文章来源: Invalid function in ruby