Python2.7左值修改的干净语法

Question

很常见的一种情况是，类似结构的类型不希望被远程版权持有者修改。在

字符串是一个基本的例子，但这是一个简单的例子，因为它是可以原谅的不可变的——Python甚至不允许对文本字符串进行方法调用之类的事情。在

问题是（在大多数语言中）我们经常有这样的东西，比如(x,y)Point类。我们偶尔想独立地改变x和{}。一、 例如，从使用的角度来看，一个Point左值应该是可变的（即使副本看不到变化）。在

但Python2.7似乎没有提供任何选项来启用赋值时自动复制。所以这意味着我们实际上必须使我们的Point类不可变，因为不经意的引用会到处被创建（通常是因为有人在将对象传递给其他人之前忘记了克隆对象）。在

不，我对无数的黑客攻击不感兴趣，这些黑客只允许在“创建对象的同时”进行变异，因为这是一个不可伸缩的弱概念。在

这些情况的逻辑结论是，我们需要我们的变异方法来实际修改左值。例如，%=支持这一点。问题是有一个更合理的语法会更好，比如使用__setattr__和/或定义set_x和{}方法，如下所示。在

class Point(object):
# Python doesn't have copy-on-assignment, so we must use an immutable
# object to avoid unintended changes by distant copyholders.

    def __init__(self, x, y, others=None):
        object.__setattr__(self, 'x', x)
        object.__setattr__(self, 'y', y)

    def __setattr__(self, name, value):
        self %= (name, value)
        return self # SHOULD modify lvalue (didn't work)

    def __repr__(self):
        return "(%d %d)" % (self.x, self.y)

    def copy(self, x=None, y=None):
        if x == None: x = self.x
        if y == None: y = self.y
        return Point(x, y)

    def __eq__ (a,b): return a.x == b.x and a.y == b.y
    def __ne__ (a,b): return a.x != b.x or  a.y != b.y
    def __add__(a,b): return Point(a.x+b.x, a.y+b.y)
    def __sub__(a,b): return Point(a.x-b.x, a.y-b.y)

    def set_x(a,b): return a.copy(x=b) # SHOULD modify lvalue (didn't work)
    def set_y(a,b): return a.copy(y=b) # SHOULD modify lvalue (didn't work)

    # This works in Python 2.7. But the syntax is awful.
    def __imod__(a,b):
        if   b[0] == 'x': return a.copy(x=b[1])
        elif b[0] == 'y': return a.copy(y=b[1])
        else:             raise AttributeError,  \
                "Point has no member '%s'" % b[0]



my_very_long_and_complicated_lvalue_expression = [Point(10,10)] * 4


# modify element 0 via "+="   -- OK
my_very_long_and_complicated_lvalue_expression[0] += Point(1,-1)

# modify element 1 via normal "__set_attr__"   -- NOT OK
my_very_long_and_complicated_lvalue_expression[1].x = 9999

# modify element 2 via normal "set_x"  -- NOT OK
my_very_long_and_complicated_lvalue_expression[2].set_x(99)

# modify element 3 via goofy "set_x"   -- OK
my_very_long_and_complicated_lvalue_expression[3]    %='x',   999


print my_very_long_and_complicated_lvalue_expression

结果是：

^{pr2}$

如您所见，+=和%=工作得很好，但是其他任何东西似乎都不起作用。当然，语言发明者已经为左值修改创建了一个基本的语法，而不仅仅局限于看起来愚蠢的运算符。我就是找不到。请帮忙。在

Answer 1

在Python中，典型的模式是在修改之前进行复制，而不是在赋值时复制。你可以用你想要的语义来实现某种数据存储，但这似乎需要大量的工作。在

Answer 2

我觉得我们已经尽职尽责地寻找已经存在的解决方案。鉴于“<；=”在某些语言中是赋值（例如，Verilog），我们可以非常直观地介绍：

value_struct_instance<<='field', value

作为Python的形式

^{pr2}$

以下是一个更新的示例，以供参考：

# Python doesn't support copy-on-assignment, so we must use an
# immutable object to avoid unintended changes by distant copyholders.
# As a consequence, the lvalue must be changed on a field update.
#
# Currently the only known syntax for updating a field on such an
# object is:
#
#      value_struct_instance<<='field', value
# 
# https://stackoverflow.com/questions/45788271/

class Point(object):

    def __init__(self, x, y, others=None):
        object.__setattr__(self, 'x', x)
        object.__setattr__(self, 'y', y)

    def __setattr__(self, name, value):
        raise AttributeError, \
            "Use \"point<<='%s', ...\" instead of \"point.%s = ...\"" \
            % (name, name)

    def __repr__(self):
        return "(%d %d)" % (self.x, self.y)

    def copy(self, x=None, y=None):
        if x == None: x = self.x
        if y == None: y = self.y
        return Point(x, y)

    def __ilshift__(a,b):
        if   b[0] == 'x': return a.copy(x=b[1])
        elif b[0] == 'y': return a.copy(y=b[1])
        else:             raise AttributeError,  \
                "Point has no member '%s'" % b[0]

    def __eq__ (a,b): return a.x == b.x and a.y == b.y
    def __ne__ (a,b): return a.x != b.x or  a.y != b.y
    def __add__(a,b): return Point(a.x+b.x, a.y+b.y)
    def __sub__(a,b): return Point(a.x-b.x, a.y-b.y)



my_very_long_and_complicated_lvalue_expression = [Point(10,10)] * 3

# modify element 0 via "+="
my_very_long_and_complicated_lvalue_expression[0] += Point(1,-1)

# modify element 1 via "<<='field'," (NEW IDIOM)
my_very_long_and_complicated_lvalue_expression[1]<<='x', 15
print my_very_long_and_complicated_lvalue_expression
# result:
# [(11 9), (15 10), (10 10)]

my_very_long_and_complicated_lvalue_expression[1]<<='y', 25
print my_very_long_and_complicated_lvalue_expression
# result:
# [(11 9), (15 25), (10 10)]

# Attempt to modify element 2 via ".field="
my_very_long_and_complicated_lvalue_expression[2].y = 25
# result:
# AttributeError: Use "point<<='y', ..." instead of "point.y = ..."