Ruby 是否能够确定键（或方法参数）是否是 proc？答案

【问题标题】：Is Ruby capable of determining whether or not a key (or method argument) is a proc?Ruby 是否能够确定键（或方法参数）是否是 proc？
【发布时间】：2012-09-07 10:40:52
【问题描述】：

在执行this code kata 时，我必须从以下方法中删除尽可能多的 if 语句：

#  returns permissions if user is set in security context
def get_user_permissions 
  user_permissions = Set.new
  if (@user != nil)
    user_permissions << :DEFAULT_PERMISSION
    if (has_cm_team_role) 
      user_permissions << :CM_TEAM_ROLE_PERMISSION
    end
    if (has_cm_invoice_view_role || has_invoice_finance_role)
      user_permissions << :CM_INVOICE_USER_PERMISSION
      user_permissions << :INVOICE_VIEW_PERMISSION
      user_permissions << :ACCESS_ALL_INVOICE_PERMISSION
    end
    if (has_invoice_finance_role) 
      user_permissions << :FINANCE_INVOICE_PERMISSION
    end
    if (has_application_access)
      user_permissions << :CM_INVOICE_USER_PERMISSION
    end
    if (has_application_access(:CM_INVOICE_ROLE)) 
      user_permissions << :CM_ANY_INVOICE_PERMISSION
    end
    if (has_application_access(:PA_INVOICE_ROLE)) 
      user_permissions << :PA_ANY_INVOICE_PERMISSION
    end
    if (has_application_access(:SDT_INVOICE_ROLE))
      user_permissions << :SDT_ANY_INVOICE_PERMISSION
    end
  end
  user_permissions
end

我的第一次尝试几乎成功，但有一些测试失败了：

def get_user_permissions 
  user_permissions = Set.new
  if (@user != nil)
    user_permissions << :DEFAULT_PERMISSION
    # add the permissions in another method
    add_permissions(user_permissions)
  end
  user_permissions
end

def add_permissions(user_permissions)
  # a hash where each key is a condition, and each value is a permission
  hash = {
    has_cm_team_role => :CM_TEAM_ROLE_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :CM_INVOICE_USER_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :INVOICE_VIEW_PERMISSION,
    has_cm_invoice_view_role || has_invoice_finance_role => :ACCESS_ALL_INVOICE_PERMISSION,
    has_invoice_finance_role => :FINANCE_INVOICE_PERMISSION,
    has_application_access => :CM_INVOICE_USER_PERMISSION,
    has_application_access(:CM_INVOICE_ROLE) => :CM_ANY_INVOICE_PERMISSION,
    has_application_access(:PA_INVOICE_ROLE) => :PA_ANY_INVOICE_PERMISSION,
    has_application_access(:SDT_INVOICE_ROLE) => :SDT_ANY_INVOICE_PERMISSION
  }

  # loop through the hash and add permissions if the key is true
  hash.each do |condition, permission|
    if (condition)
      user_permissions << permission
    end
  end

这种方法的问题是三个 OR 语句（在我的哈希的第二个、第三个和第四个键中）不是由 each 评估的。所以我通过使用 Procs 来解决这个问题，如下所示：

def add_permissions(user_permissions)
  hash = {
    Proc.new{has_cm_team_role} => :CM_TEAM_ROLE_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :INVOICE_VIEW_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :ACCESS_ALL_INVOICE_PERMISSION,
    Proc.new{has_invoice_finance_role} => :FINANCE_INVOICE_PERMISSION,
    Proc.new{has_application_access} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_application_access(:CM_INVOICE_ROLE)} => :CM_ANY_INVOICE_PERMISSION,
    Proc.new{has_application_access(:PA_INVOICE_ROLE)} => :PA_ANY_INVOICE_PERMISSION,
    Proc.new{has_application_access(:SDT_INVOICE_ROLE)} => :SDT_ANY_INVOICE_PERMISSION
  }
  hash.each do |condition, permission|
    if (condition.call)
      user_permissions << permission
    end
  end
end

好的，这行得通，测试全部通过，但我将所有键转换为 Procs，以便其中 3 个正确评估。我宁愿只在需要时使用 Procs，即第二个、第三个和第四个键，如下所示：

def add_permissions(user_permissions)
  hash = {
    # just use the method name unless a Proc is required
    has_cm_team_role => :CM_TEAM_ROLE_PERMISSION,
    # use a Proc here, so that the OR is evaluated later
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :CM_INVOICE_USER_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :INVOICE_VIEW_PERMISSION,
    Proc.new{has_cm_invoice_view_role || has_invoice_finance_role} => :ACCESS_ALL_INVOICE_PERMISSION,
    has_invoice_finance_role => :FINANCE_INVOICE_PERMISSION,
    has_application_access => :CM_INVOICE_USER_PERMISSION,
    has_application_access(:CM_INVOICE_ROLE) => :CM_ANY_INVOICE_PERMISSION,
    has_application_access(:PA_INVOICE_ROLE) => :PA_ANY_INVOICE_PERMISSION,
    has_application_access(:SDT_INVOICE_ROLE) => :SDT_ANY_INVOICE_PERMISSION
  }

然后用某种方法来判断key是否是Proc，如果是，就调用它，如果不是，就像对待其他key一样处理条件：

  hash.each do |condition, permission|
    if condition.is_proc?
      if (condition.call)
        user_permissions << permission
      end
    elsif condition
      user_permissions << permission
    end
  end
end

有什么建议吗？有什么比我尝试这样做更好的想法吗？通过这样做，我是否让它变得更加循环复杂？我应该坚持使用我所有的键都是 Procs 的工作解决方案吗？

【问题讨论】：

标签： ruby proc

【解决方案1】：

condition.is_a? Proc满足你的条件吗？

【讨论】：

【解决方案2】：

与 proc 问题无关。我从未做过代码 kata，但我认为你可以改进的地方很少。

首先，将条件作为哈希键对我来说是个坏主意。您将有许多键重复，只有 false 和 true。
我会以另一种方式进行，并将您的权限设置为键，值将是 true 或 false。

您也有 3 次完全相同的说明，每次只更改 2 个字母。

if (has_application_access(:CM_INVOICE_ROLE)) 
  user_permissions << :CM_ANY_INVOICE_PERMISSION
end
if (has_application_access(:PA_INVOICE_ROLE)) 
  user_permissions << :PA_ANY_INVOICE_PERMISSION
end
if (has_application_access(:SDT_INVOICE_ROLE))
  user_permissions << :SDT_ANY_INVOICE_PERMISSION
end

可以简化为

%(cm pa sdt).each do |key|
    user_permissions << :"#{key}_ANY_INVOICE_PERMISSION" if has_application_access(:"#{key}_INVOICE_ROLE")
end

我尝试自己这样做并得到了两个结果。一种使用哈希，一种尽可能多地切割。

# Hash version
def get_user_permissions
  return Set.new if !!@user

  user_permissions_hash = {:DEFAULT_PERMISSION => true,
                           :CM_TEAM_ROLE_PERMISSION => has_cm_team_role,
                           :CM_INVOICE_USER_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role || has_application_access,
                           :INVOICE_VIEW_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role,
                           :ACCESS_ALL_INVOICE_PERMISSION => has_cm_invoice_view_role || has_invoice_finance_role,
                           :FINANCE_INVOICE_PERMISSION => has_invoice_finance_role
  }
  %(cm pa sdt).each do |key|
    user_permissions_hash[:"#{key}_ANY_INVOICE_PERMISSION"] = has_application_access(:"#{key}_INVOICE_ROLE")
  end

  return user_permissions_hash.map {|k, v| k if v}.compact.to_set
end

# Normal version
def get_user_permissions
  return (user_permissions = Set.new) if !!@user

  user_permissions << :DEFAULT_PERMISSION
  user_permissions << :CM_TEAM_ROLE_PERMISSION if has_cm_team_role
  user_permissions << :CM_INVOICE_USER_PERMISSION if has_cm_invoice_view_role || has_invoice_finance_role || has_application_access
  if (has_cm_invoice_view_role || has_invoice_finance_role)
    user_permissions << :INVOICE_VIEW_PERMISSION
    user_permissions << :ACCESS_ALL_INVOICE_PERMISSION
  end
  user_permissions << :FINANCE_INVOICE_PERMISSION if has_invoice_finance_role
  %(cm pa sdt).each do |key|
    user_permissions << :"#{key}_ANY_INVOICE_PERMISSION" if has_application_access(:"#{key}_INVOICE_ROLE")
  end

  user_permissions
end

【讨论】：

嘿，非常感谢您的建议，尤其是关于切换键和值的提示，这很有意义。代码也不错。
PS - 我必须从第一个 return 语句中删除一个刘海（即return Set.new if !@user），我还必须将%(cm pa sdt) 中的字符串大写，但这很好用。再次感谢。