Python decorator@func（）.attribute语法

Question

我试图在这里找到答案，但没找到。在

@obj.func # works
@obj.func(**kwargs)  #works
@obj.func1(**kwargs).func2   #-> syntax error 

我不明白为什么第三种形式是SyntaxError，对我来说似乎没有违反任何python语法，而且用户想要做什么我很清楚（见下面的示例）。在

我查看了decorator实现的pep 0318，但没有找到任何答案。在

下面是一个使用示例：

^{pr2}$

可以使用ItemFunc作为装饰器：

@ItemFunc
def plot(**kwargs):
    pass

redcross = plot.copy(color="red", marker="+")
@redcross.caller
def plot_data1(**kwargs):
    pass

bluecross = redcross.copy(color="blue")
@bluecross.caller
def plot_data2(**kwargs):
    pass

但为什么禁止使用以下“快捷语法”：

@redcross.copy(color="blue").caller
def plot_data2(**kwargs):
    pass

但我可以：

@redcross.copy_and_decorate(color="blue")
def plot_data2(**kwargs):
    pass         

第一种形式看起来更好，至少我更了解背后的意图。在

Answer 1

Function definitions grammar不允许使用更多的虚线名称调用；语法仅限于虚线名称和结尾处的可选调用：

decorated      ::=  decorators (classdef | funcdef)
decorators     ::=  decorator+
decorator      ::=  "@" dotted_name ["(" [argument_list [","]] ")"] NEWLINE
funcdef        ::=  "def" funcname "(" [parameter_list] ")" ":" suite
dotted_name    ::=  identifier ("." identifier)*

注意，这不是一个完整的表达式，而是一个非常有限的子集。在

这与政治公众人物计划相呼应，该计划指出：

The decorator statement is limited in what it can accept arbitrary expressions will not work. Guido preferred this because of a gut feeling [17] .

以及

The rationale for having a function that returns a decorator is that the part after the @ sign can be considered to be an expression (though syntactically restricted to just a function), and whatever that expression returns is called. See declaration arguments [16] .

强调我的。在

理由是Guido feels there isn't a real use case for allowing more：

So while it would be quite easy to change the syntax to @test in the future, I'd like to stick with the more restricted form unless a real use case is presented where allowing @test would increase readability. (@foo().bar() doesn't count because I don't expect you'll ever need that).

您必须说服Guido和其他核心开发人员，您的案例是一个适当的用例，值得取消这些限制！在