[英]Instance methods defined between a class and a certain ancestor of it
假设A
类是B
类的后代。 获取B
定义的A
的所有实例方法的列表(符号数组,顺序无关紧要)的最佳方法是什么,即在B
或其任何后代类中定义的A
实例方法? 示例如下。 类层次结构如下:
class C; def c; end end
class B < C; def b; end end
class D < B; def d; end end
class A < D; def a; end end
各种类中A
的实例方法是:
A.instance_methods_within(Kernel) # => (Same as A.instance_methods)
A.instance_methods_within(C) # => [:a, :d, :b, :c]
A.instance_methods_within(B) # => [:a, :d, :b]
A.instance_methods_within(D) # => [:a, :d]
A.instance_methods_within(A) # => [:a] (Same as A.instance_methods(false))
这是我自己的解决方案。
class Class
def instance_methods_within klass
ancestors
.select{|k| k <= klass}
.inject([]){|a, k| a.concat(k.instance_methods(false))}
.uniq
end
end
Marc-AndréLafortune建议:
class Class
def instance_methods_within klass
ancestors
.select{|k| k <= klass}
.flat_map{|k| k.instance_methods(false)}
.uniq
end
end
我相信你在寻找这个:
class Class
def instance_methods_within(klass)
return self.instance_methods if klass == Object
methods = []
this = self
while this != nil
methods << this.instance_methods(false)
break if this == klass
this = this.superclass
end
return methods.flatten
end
end
class C; def c; end end
class B < C; def b; end end
class D < B; def d; end end
class A < D; def a; end end
A.instance_methods_within(Object) # => (Same as A.instance_methods)
A.instance_methods_within(C) # => [:a, :d, :b, :c]
A.instance_methods_within(B) # => [:a, :d, :b]
A.instance_methods_within(D) # => [:a, :d]
A.instance_methods_within(A) # => [:a] (Same as A.instance_methods(false))
我添加了Object
as Object.instance_methods(false)
yield []
的特殊情况,因此在这种情况下向上遍历的循环不会很好。 更新以将instance_methods_within
类定义为Class
的实例方法。
我想提出这个问题: -
class Class
def instance_methods_within(klass = nil)
imethods = instance_methods
own_imethods = instance_methods(false)
return imethods if [Object, Kernel, nil].include?(klass) || [BasicObject, nil].include?(klass.superclass)
sc_imethods = klass.superclass.instance_methods
own_imethods | (imethods - sc_imethods)
end
end
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