所以我有各种哈希值,它们并不总是具有相同的键/值对 . 我想要做的是能够合并散列,但是如果它们不存在于该散列中,则添加空键/值对,但在其他散列中也是如此 . 这很难解释,但这可能会更好地解释它:
t1 = Merger.new({"a" => "1"})
puts t1.merge({:b => 2})
# => {"a" => "1", :b => 2}
t2 = Merger.new({"a" => "1", :b => 2})
puts t2.merge({:c => "3"})
# => {"a" => "1", :b => 2, :c => "3"}
t3 = Merger.new({"a" => "1", "b" => 2})
puts t3.merge
# => {"a" => "1", "b" => 2}
t4 = Merger.new
puts t4.merge({:a => 1})
# => {:a => 1}
t5 = Merger.new
puts t4.merge
# => {}
Merger
类实现:
class Merger
alias AnyHash = Hash(Symbol | String, Int32 | String) |
Hash(Symbol, Int32 | String) |
Hash(String, Int32 | String) |
Hash(String, String) |
Hash(String, Int32) |
Hash(Symbol, String) |
Hash(Symbol, Int32)
def initialize(params : AnyHash? = nil)
@params = params
end
def merge(other = {} of String => String)
@params.try do |params|
other = params.merge(other)
end
return other
end
end
https://play.crystal-lang.org/#/r/3oeh
从字面上看,我应该创建具有键/值对的所有可能组合的联合类型 . 否则,将给出编译时错误 .
是否有更优雅的方式使其工作?
1 回答
Hash#merge
已经为您找出了正确的泛型类型参数 .如果我理解正确的话,唯一的问题是将其存储在实例变量中 . 这可以使
Merger
成为一个泛型类,其中类型是从构造函数参数自动推断出来的: