Question

我正在尝试编写helper_method类型的功能。这样做我遇到了这种奇怪的行为：

irb(main):001:0> def a; end
=> :a
irb(main):002:0> b = {}
=> {}
irb(main):003:0> a(**{})
=> nil
irb(main):004:0> a(**b)
ArgumentError: wrong number of arguments (given 1, expected 0)
  from (irb):1:in `a'
  from (irb):4
  from /home/vagrant/.rbenv/versions/2.3.1/bin/irb:11:in `<main>'

方法a不带参数。通过splatting一个空哈希来调用它可以工作，但是这个哈希存储在一个它失败的变量中。这看起来像一个合法的错误。有人有什么想法吗？

Answer 1

一些半知情的猜测：在其中使用doublesplat的方法调用会将doublesplat的参数与任何现有的关键字参数合并，并将其作为散列放入最后一个参数中。使用文字空哈希，编译器可以看到没有关键字，并且可以跳过创建哈希：

puts RubyVM::InstructionSequence.compile("b = {}; a(**{})").disasm
== disasm: #<ISeq:<compiled>@<compiled>>================================
local table (size: 2, argc: 0 [opts: 0, rest: -1, post: 0, block: -1, kw: -1@-1, kwrest: -1])
[ 2] b          
0000 trace            1                                               (   1)
0002 newhash          0
0004 setlocal_OP__WC__0 2
0006 putself          
0007 opt_send_without_block <callinfo!mid:a, argc:0, FCALL|ARGS_SIMPLE>, <callcache>
0010 leave

只要关键字总数可预测为零，您就会得到相同的结果。因此，这两个编译相同：

a()
a(**{})

使用变量哈希（恰好是空的），不能做出这个假设，并且总是调用合并，产生哈希参数：

puts RubyVM::InstructionSequence.compile("b = {}; a(**b)").disasm
== disasm: #<ISeq:<compiled>@<compiled>>================================
local table (size: 2, argc: 0 [opts: 0, rest: -1, post: 0, block: -1, kw: -1@-1, kwrest: -1])
[ 2] b          
0000 trace            1                                               (   1)
0002 newhash          0
0004 setlocal_OP__WC__0 2
0006 putself          
0007 putspecialobject 1
0009 getlocal_OP__WC__0 2
0011 opt_send_without_block <callinfo!mid:core#hash_merge_kwd, argc:1, ARGS_SIMPLE>, <callcache>
0014 opt_send_without_block <callinfo!mid:dup, argc:0, ARGS_SIMPLE>, <callcache>
0017 opt_send_without_block <callinfo!mid:a, argc:1, FCALL|ARGS_SIMPLE>, <callcache>
0020 leave

所以我猜2.2.2添加了一些优化代码，它们看到了**{}的无效，并且跳过了生成额外的代码。

如果您定义方法以始终收集其余关键字，则不会中断，因为即使发件人未传递哈希，也会创建哈希：

def a(**x); p x; end
a()         # works, x is {} - hash supplied by VM at receiver
a(**b)      # works, x is {} - hash supplied by sender as b.merge({})
a(**{})     # works, x is {} - hash supplied by VM at receiver, **{} ignored
a(b, **{})  # works, x is {} - hash supplied by sender, **{} ignored
a({}, **{}) # works, x is {} - hash supplied by sender, **{} ignored
a(b, **b)   # works, x is {} - hash supplied by sender as b.merge(b)
a({}, **b)  # works, x is {} - hash supplied by sender as b.merge({})

为不带参数的方法splatting哈希时的不同行为

1 个答案: