重载 __eq__ 返回自定义对象
我正在用Python写一个领域特定语言（DSL），想要重载一些运算符，这样我就可以轻松地写出我的DSL表达式。比如，我想写 Var("a") + Var("b")，然后得到 Add(Var("a"), Var("b")) 这样的表示。为此，我重载了 __add__ 方法，这个方法对我来说效果很好。
不过，我也想重载 __eq__ 方法，达到类似的效果：我希望写 Var("a") == Var("b")，然后得到 Eq(Var("a"), Var("b")) 这样的表示。通过重载 __eq__ 方法并返回一个 Eq 的实例，我实现了这个目标。但是，重载 __eq__ 方法显然会干扰Python的标准行为，比如 Var("b") in [Var("a")] 会返回 True。
有没有办法让我实现这个目标，也就是能够写 Var("a") == Var("b") 并得到 Eq(Var("a"), Var("b"))，同时还能写 if Var("a") == Var("b"): blablabla 或者把表达式放进内置的容器里等等？
编辑
我尝试实现了 Eq 类的 __bool__ 方法，似乎有效（见下面的代码）。我是不是漏掉了什么，或者这是一个可行的解决方案？
class Expr:
def __add__(self, other):
return Add(self, other)
def __eq__(self, other):
return Eq(self, other)
def __repr__(self):
return str(self)
def __add__(self, other):
return Add(self, other)
def __ne__(self, other):
return Neq(self, other)
class Var(Expr):
def __init__(self, name):
self.name = name
def __str__(self):
return "Var(" + str(self.name) + ")"
def equals(self, other):
if type(self) is type(other):
return self.name == other.name
else:
return False
def __hash__(self):
return 17 + 23 * hash(self.name)
class Add(Expr):
def __init__(self, left, right):
self.left = left
self.right = right
def __str__(self):
return "Add(" + str(self.left) + ", " + str(self.right) + ")"
def equals(self, other):
if type(self) is type(other):
return ( ( self.left.equals(other.left) and
self.right.equals(other.right) ) or
( self.left.equals(other.right) and
self.right.equals(other.left) ) )
else:
return False
def __hash__(self):
return (17 + 23 * hash("+") +
23 * 23 * hash(self.left) + 23 * 23 * hash(self.right))
class Eq(Expr):
def __init__(self, left, right):
self.left = left
self.right = right
def __str__(self):
return "Eq(" + str(self.left) + ", " + str(self.right) + ")"
def equals(self, other):
if type(self) is type(other):
return ( ( self.left.equals(other.left) and
self.right.equals(other.right) ) or
( self.left.equals(other.right) and
self.right.equals(other.left) ) )
else:
return False
def __bool__(self):
return self.left.equals(self.right)
def __hash__(self):
return (17 + 23 * hash("==") +
23 * 23 * hash(self.left) + 23 * 23 * hash(self.right))
class Neq(Expr):
def __init__(self, left, right):
self.left = left
self.right = right
def __str__(self):
return "Neq(" + str(self.left) + ", " + str(self.right) + ")"
def equals(self, other):
if type(self) is type(other):
return ( ( not self.left.equals(other.left) or
not self.right.equals(other.right) ) and
( not self.left.equals(other.right) or
not self.right.equals(other.left) ) )
else:
return False
def __bool__(self):
return not self.left.equals(self.right)
def __hash__(self):
return (17 + 23 * hash("!=") +
23 * 23 * hash(self.left) + 23 * 23 * hash(self.right))
a = Var("a")
aa = Var("a")
b = Var("b")
c = Var("c")
print("a + b", "=>", a + b) # a + b => Add(Var(a), Var(b))
print("a == b", "=>", a == b) # a == b => Eq(Var(a), Var(b))
print("a != b", "=>", a != b) # a != b => Neq(Var(a), Var(b))
print("a if a == b else b", "=>", a if a == b else b)
# a if a == b else b => Var(b)
print("a if a == aa else b", "=>", a if a == aa else b)
# a if a == aa else b => Var(a)
l = [a, a+b]
print("l", "=>", l) # l => [Var(a), Add(Var(a), Var(b))]
print("b in l", "=>", b in l) # b in l => False
print("a in l", "=>", a in l) # a in l => True
print("aa in l", "=>", aa in l) # aa in l => True
print("a+b in l", "=>", a+b in l) # a+b in l => True
print("b+a in l", "=>", b+a in l) # b+a in l => True
print("a+c in l", "=>", a+c in l) # a+c in l => False
if a == b:
print("a == b is True")
else:
print("a == b is False") # a == b is False
if a == aa:
print("a == aa is True") # a == aa is True
else:
print("a == aa is False")
if a != b:
print("a != b is True") # a != b is True
else:
print("a != b is False")
if a != aa:
print("a != aa is True")
else:
print("a != aa is False") # a != aa is False
if a == b or a == aa:
print("a == b or a == aa is True") # a == b or a == aa is True
else:
print("a == b or a == aa is False")
if a == aa and a == b:
print("a == aa and a == b is True")
else:
print("a == aa and a == b is False") # a == aa and a == b is False
if not a == aa:
print("not a == aa is True")
else:
print("not a == aa is False") # not a == aa is False
if not a == b:
print("not a == b is True") # not a == b is True
else:
print("not a == b is False")
if a == 3:
print("a == 3 is True")
else:
print("a == 3 is False") # a == 3 is False
if a != 3:
print("a != 3 is True") # a != 3 is True
else:
print("a != 3 is False")
if 3 == a:
print("3 == a is True")
else:
print("3 == a is False") # 3 == a is False
if 3 != a:
print("3 != a is True") # 3 != a is True
else:
print("3 != a is False")
if a == 'a':
print("a == 'a' is True")
else:
print("a == 'a' is False") # a == 'a' is False
if a != 'a':
print("a != 'a' is True") # a != 'a' is True
else:
print("a != 'a' is False")
if 'a' == a:
print("'a' == a is True")
else:
print("'a' == a is False") # 'a' == a is False
if 'a' != a:
print("'a' != a is True") # 'a' != a is True
else:
print("'a' != a is False")
s = {a}
print("s", "=>", s) # s => {Var(a)}
print("a in s", "=>", a in s) # a in s => True
print("b in s", "=>", b in s) # b in s => False
print("aa in s", "=>", aa in s) # aa in s => True
d = {a: 1, b: 2}
print("d", "=>", d) # d => {Var(b): 2, Var(a): 1}
print("d[a]", "=>", d[a]) # d[a] => 1
print("d[b]", "=>", d[b]) # d[b] => 2
print("c in d", "=>", c in d) # c in d => False
print("aa in d", "=>", aa in d) # aa in d => True
print("d[aa]", "=>", d[aa]) # d[aa] => 1
1 个回答
2
不可以，你必须选择一种行为。.__eq__()这个方法的使用场景是无法可靠地检测到的。
如果你需要同时具备这两种行为，那你就得使用其他的运算符或者方法来表示你想要的功能。