为什么 `x += a + d[x]` 工作正常但 `x +=a; x += d[x]` 失败？

Question

我最近重构了这段代码，但我弄坏了它：

            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])

经过我的重构，为了让行更短，它看起来像这样：

            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])

变量的类型是：

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

这种重构的不良影响是我得到这样的字符串：

"\nAnnounces: \nAnnounces: \nAnnounces: "

这可能是什么原因？

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这是整个函数及其输出：

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("}")

并且输出：

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=?
"?" -> "?";
}

Answer 1

正如用户 lurker 在他的评论中所说，这是因为第一个表达式更改了 from_mac 的内容，然后字典中的查找失败。一种解决方案是引入一个仅用于显示的新变量。

为什么 `x += a + d[x]` 工作正常但 `x +=a; x += d[x]` 失败？

Why `x += a + d[x]` works fine but `x +=a; x += d[x]` fails?

python

refactoring

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