最近我重构这个片段，我把它弄坏了：为什么`x + = a + d [x]`可以正常工作，但`x + = a; x + = d [x]`失败？

  if from_mac in self.announces:          
       from_mac += '\\nAnnounces: ' + ',\\n'.join(self.announces[from_mac]) 
      if to_mac in self.announces:          
       to_mac += '\\nAnnounces: ' + ',\\n'.join(self.announces[to_mac])

我在重构之后，它看起来像这样使线较短：

  if from_mac in self.announces:          
       from_mac += '\\nAnnounces: ' 
       from_mac += ',\\n'.join(self.announces[from_mac]) 
      if to_mac in self.announces:          
       to_mac += '\\nAnnounces: ' 
       to_mac += ',\\n'.join(self.announces[to_mac])

的类型变量是：

to_mac, from_mac -> string 
self.announces = defaultdict(list) # of strings

此重构的不良影响是我得到像这样的字符串：

"\nAnnounces: \nAnnounces: \nAnnounces: "

可能是什么原因呢？

语境

这里的整体功能和它的输出：

def print_report(self, skip_broadcast=False):         
    """                   
    Prints out a DOT file based on the gathered information.     
    """                   
    sys.stderr.write("%s\n" % repr(self.announces))        
    print("strict digraph {")             
    for from_mac in self.seen:             
     sys.stderr.write("k1=%s\n" % from_mac)         
     for to_mac in self.seen[from_mac]:          
      sys.stderr.write("k2=%s\n" % to_mac)        
      if skip_broadcast and (from_mac == '?' or to_mac == '?'):   
       continue              
      if from_mac in self.announces:          
       from_mac += '\\nAnnounces: '         
       from_mac += ',\\n'.join(self.announces[from_mac])    
      if to_mac in self.announces:          
       to_mac += '\\nAnnounces: '          
       to_mac += ',\\n'.join(self.announces[to_mac])     
      print('"%s" -> "%s";' % (from_mac, to_mac))       
    print("}")

和输出：

defaultdict(<type 'list'>, {'Cisco-Li_99:13:54\\n(58:6d:8f:99:13:54)': ['FajnaSiec']}) 
strict digraph { 
k1=Cisco-Li_99:13:54\n(58:6d:8f:99:13:54) 
k2=IPv6mcast_01\n(33:33:00:00:00:01) 
"Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: " -> "IPv6mcast_01\n(33:33:00:00:00:01)"; 
k2=IPv4mcast_01\n(01:00:5e:00:00:01) 
"Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: \nAnnounces: " -> "IPv4mcast_01\n(01:00:5e:00:00:01)"; 
k2=Tp-LinkT_20:74:8b\n(e8:94:f6:20:74:8b) 
"Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: \nAnnounces: \nAnnounces: " -> "Tp-LinkT_20:74:8b\n(e8:94:f6:20:74:8b)"; 
k2=? 
"Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: \nAnnounces: \nAnnounces: \nAnnounces: " -> "?"; 
k1=SamsungE_05:50:0e\n(00:e3:b2:05:50:0e) 
k2=Cisco-Li_99:13:54\n(58:6d:8f:99:13:54) 
"SamsungE_05:50:0e\n(00:e3:b2:05:50:0e)" -> "Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: "; 
k1=? 
k2=Cisco-Li_99:13:54\n(58:6d:8f:99:13:54) 
"?" -> "Cisco-Li_99:13:54\n(58:6d:8f:99:13:54)\nAnnounces: "; 
k2=SamsungE_05:50:0e\n(00:e3:b2:05:50:0e) 
"?" -> "SamsungE_05:50:0e\n(00:e3:b2:05:50:0e)"; 
k2=? 
"?" -> "?"; 
}

来源

2015-05-03 d33tah

在您的重构示例中，右手表达式中'from_mac'的值与您原始代码中的* *不同。换句话说，这不仅仅是'x + = a + b'的一个简单情况，而且这里'b'取决于'x'的值。所以'x + = a + f（x）'与* x + = a'相同，后面是'x + = f（x）'，因为第二种情况下'x'的值是*不同的*。 – lurker

啊废话！愚蠢的错误。谢谢！ – d33tah

由于用户lurker在他的评论中说，这是因为第一个表达式的变化from_mac是什么，然后字典中的查找失败。解决方案是引入一个新的变量，仅用于显示。

来源

2015-05-03 11:02:24 d33tah

为什么`x + = a + d [x]`可以正常工作，但`x + = a; x + = d [x]`失败？

语境

回答

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