Python3求余容易犯错的点

想获得IP地址递增的字符串

用Python2，风平浪静

$ cat python2_ip.py

for i in range(10):
    print '11.0.{0}.{1}/32'.format(i/256,i%256)
$
$ python  python2_ip.py
11.0.0.0/32
11.0.0.1/32
11.0.0.2/32
11.0.0.3/32
11.0.0.4/32
11.0.0.5/32
11.0.0.6/32
11.0.0.7/32
11.0.0.8/32
11.0.0.9/32
           

换成Python3，怎么出这种结果了？

$ cat python3_ip.py 

for i in range(10):
    print('11.0.{0}.{1}/32'.format(i/256,i%256))
$
$ python3 python3_ip.py 
11.0.0.0.0/32
11.0.0.00390625.1/32
11.0.0.0078125.2/32
11.0.0.01171875.3/32
11.0.0.015625.4/32
11.0.0.01953125.5/32
11.0.0.0234375.6/32
11.0.0.02734375.7/32
11.0.0.03125.8/32
11.0.0.03515625.9/32
           

问题出在i/256！

在python2中，"整数/整数"得到的是商；

在python3中，"整数/整数"得到的是

精确值

，"整数//整数"得到的是商，注意是两个//

$ cat python3_ip.py 

for i in range(10):
    print('11.0.{0}.{1}/32'.format(i/256,i%256))
$
$ python3 python3_ip.py 
11.0.0.0/32
11.0.0.1/32
11.0.0.2/32
11.0.0.3/32
11.0.0.4/32
11.0.0.5/32
11.0.0.6/32
11.0.0.7/32
11.0.0.8/32
11.0.0.9/32