Bitwise XOR运算符不在python中工作

I am trying to do this :

我想这样做：

print 17593028247552 ^ 909522486

打印17593028247552 ^ 909522486

The result must be 67450422(as is in javascript) but I am getting 17592253494838L

结果必须是67450422（在javascript中），但我得到17592253494838L

Please help me ! Thanks in advance :)

请帮帮我！提前致谢：）

2 个解决方案

#1

Python 2's long type (or int in Python 3) can grow as large as they need to, but it looks like you want a 32 bit result.

Python 2的long类型（或Python 3中的int）可以根据需要增长，但看起来你想要32位结果。

You just need to mask the result so you only get the low 32 bits of the result. And I guess that since you're on Python 2 you should also convert it from long to int.

您只需要屏蔽结果，这样您只能获得结果的低32位。我想，既然你在Python 2上，你也应该将它从long转换为int。

>>> int((17593028247552 ^ 909522486 ) & 0xffffffff)
67450422

#2

I just used Windows calculator:

我只使用Windows计算器：

100000000000000110010001100110000000000000000  = 17593028247552d
000000000000000110110001101100011011000110110  =      909522486d
100000000000000000100000001010011011000110110  = 17592253494838d

Looks like JavaScript is wrong. It's probably narrowing to 32 bit or something before doing XOR. Python is arbitrary precision on integers.

看起来像JavaScript是错误的。在进行XOR之前，它可能会缩小到32位。 Python是整数的任意精度。

Edit: As the other answer points out, if you're still using Python 2 (don't do that!) only long is arbitrary precision; the precision of int can vary but it's usually 64 bit. In Python 3 there is only one integer type and it's always arbitrary precision.

编辑：正如另一个答案所指出的那样，如果你还在使用Python 2（不要这样做！），那么长期就是任意精度; int的精度可以变化，但通常是64位。在Python 3中，只有一个整数类型，它始终是任意精度。

#1