20讲课后题

zyz2017

如果将：

1. if "\n" in x[i-3:i+4]:

复制代码

换为：

1. if "\n"or" "or"\r" in x[i-3:i+4]:

复制代码

将会什么都不显示，为什么？

1. def find():
   
2.     x='''ACFiCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
   
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105. qKWGc'''
     
106.     length=len(x)
     
107.     for i in range (3,length-3):
     
108.         if x[i].islower():
     
109.             if "\n" in x[i-3:i+4]:
     
110.                 continue
     
111.             if x[i-3:i].isupper()and x[i+1:i+4].isupper():
     
112.                
     
113.                 if not x[i-4].isupper()and not x[i+4].isupper():
     
114.                         print(x[i])
     
115. 
     

复制代码

BngThea

是不是优先级的问题，把in前面的一串括起来试试

铁棍阿童木

逻辑运算符 or 的运算优先级要大于成员操作符 in

因此 if "\n"or" "or"\r" in x[i-3:i+4]: 这一条件判断语句，会先把 "\n"or" "or"\r" 运算完成才会去进行成员运算，所以，从本质上来讲，这句判断语句与if "\n" in x[i-3:i+4]:是等价的。因为“\n”or ""返回“\n”，而“\n”or “\r”返回“\n”。

所以，程序并不会判断“”和“\r”是否在x里面。

zyz2017

铁棍阿童木 发表于 2017-10-20 11:20
逻辑运算符 or 的运算优先级要大于成员操作符 in

因此 if "\n"or" "or"\r" in x: 这一条件判断语句，会 ...

如果这样说加“”or“\r”和不加是一样的，但是不加是正常的。加了后什么都不会返回了

zyz2017

BngThea 发表于 2017-10-20 09:04
是不是优先级的问题，把in前面的一串括起来试试

加括号会返回但是是按得or最前面的第一个返回的。。。