我如何使用预递增/递减操作符(++,——),就像在c++ ?
为什么++计数运行,但不改变变量的值?
我如何使用预递增/递减操作符(++,——),就像在c++ ?
为什么++计数运行,但不改变变量的值?
当前回答
一个直截了当的解决办法
c = 0
c = (lambda c_plusplus: plusplus+1)(c)
print(c)
1
不再打字
c = c + 1
你也可以直接写 c++ 完成所有代码,然后搜索/替换“c++”,替换为“c=c+1”。只要确保正则表达式搜索是关闭的。
其他回答
在Python中,表达式和语句之间有严格的区别 与Common Lisp、Scheme或 Ruby。
维基百科
因此,通过引入这样的操作符,可以打破表达式/语句的分割。
和你不能写作的原因一样
if x = 0:
y = 1
就像在其他一些不保留这种区别的语言中一样。
扩展Henry的回答,我实验性地实现了一个语法糖库,实现了++:hdytto。
使用方法简单。从PyPI安装后,放置sitecustomize.py:
from hdytto import register_hdytto
register_hdytto()
在项目目录中。然后,使main.py:
# coding: hdytto
a = 5
print(a++)
print(++a)
b = 10 - --a
print(b--)
并通过PYTHONPATH=运行它。python main.py。输出将是
5
7
4
在解码脚本文件时,Hdytto将++替换为((a:=a+1)-1),因此它可以工作。
++不是运算符。它是两个+运算符。+运算符是单位运算符,它什么也不做。(澄清:+和-一元操作符仅适用于数字,但我假定您不会期望假设的++操作符适用于字符串。)
++count
解析为
+(+count)
翻译过来就是
count
你必须使用稍微长一点的+=运算符来做你想做的事情:
count += 1
我怀疑省略++和——操作符是为了保持一致性和简单性。我不知道Guido van Rossum为这个决定给出的确切理由,但我可以想象出一些理由:
Simpler parsing. Technically, parsing ++count is ambiguous, as it could be +, +, count (two unary + operators) just as easily as it could be ++, count (one unary ++ operator). It's not a significant syntactic ambiguity, but it does exist. Simpler language. ++ is nothing more than a synonym for += 1. It was a shorthand invented because C compilers were stupid and didn't know how to optimize a += 1 into the inc instruction most computers have. In this day of optimizing compilers and bytecode interpreted languages, adding operators to a language to allow programmers to optimize their code is usually frowned upon, especially in a language like Python that is designed to be consistent and readable. Confusing side-effects. One common newbie error in languages with ++ operators is mixing up the differences (both in precedence and in return value) between the pre- and post-increment/decrement operators, and Python likes to eliminate language "gotcha"-s. The precedence issues of pre-/post-increment in C are pretty hairy, and incredibly easy to mess up.
Python没有这些操作符,但如果你真的需要它们,你可以编写一个具有相同功能的函数。
def PreIncrement(name, local={}):
#Equivalent to ++name
if name in local:
local[name]+=1
return local[name]
globals()[name]+=1
return globals()[name]
def PostIncrement(name, local={}):
#Equivalent to name++
if name in local:
local[name]+=1
return local[name]-1
globals()[name]+=1
return globals()[name]-1
用法:
x = 1
y = PreIncrement('x') #y and x are both 2
a = 1
b = PostIncrement('a') #b is 1 and a is 2
在函数内部,如果你想改变局部变量,你必须添加locals()作为第二个参数,否则它将尝试改变全局变量。
x = 1
def test():
x = 10
y = PreIncrement('x') #y will be 2, local x will be still 10 and global x will be changed to 2
z = PreIncrement('x', locals()) #z will be 11, local x will be 11 and global x will be unaltered
test()
还有这些函数,你可以做:
x = 1
print(PreIncrement('x')) #print(x+=1) is illegal!
但在我看来,以下方法更清晰:
x = 1
x+=1
print(x)
递减运算符:
def PreDecrement(name, local={}):
#Equivalent to --name
if name in local:
local[name]-=1
return local[name]
globals()[name]-=1
return globals()[name]
def PostDecrement(name, local={}):
#Equivalent to name--
if name in local:
local[name]-=1
return local[name]+1
globals()[name]-=1
return globals()[name]+1
我使用这些函数在我的模块翻译javascript到python。
一个直截了当的解决办法
c = 0
c = (lambda c_plusplus: plusplus+1)(c)
print(c)
1
不再打字
c = c + 1
你也可以直接写 c++ 完成所有代码,然后搜索/替换“c++”,替换为“c=c+1”。只要确保正则表达式搜索是关闭的。