如何让位旋转函数接受任何位大小？
我有从其他代码中获得的这两个函数
def ROR(x, n):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (32 - n))
def ROL(x, n):
return ROR(x, 32 - n)
，我想在需要 16 位旋转的程序中使用它们。然而，还有其他函数需要 32 位旋转，所以我想在方程中保留 32 位，所以我得到：
def ROR(x, n, bits = 32):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (bits - n))
def ROL(x, n, bits = 32):
return ROR(x, bits - n)
然而，当我测试这个设置时，答案是错误的。 时，值会正确显示，我该如何解决这个问题？
def ROR(x, n):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (16 - n))
def ROL(x, n,bits):
return ROR(x, 16 - n)
然而，当代码发生
i have these 2 functions i got from some other code
def ROR(x, n):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (32 - n))
def ROL(x, n):
return ROR(x, 32 - n)
and i wanted to use them in a program, where 16 bit rotations are required. however, there are also other functions that require 32 bit rotations, so i wanted to leave the 32 in the equation, so i got:
def ROR(x, n, bits = 32):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (bits - n))
def ROL(x, n, bits = 32):
return ROR(x, bits - n)
however, the answers came out wrong when i tested this set out. yet, the values came out correctly when the code is
def ROR(x, n):
mask = (2L**n) - 1
mask_bits = x & mask
return (x >> n) | (mask_bits << (16 - n))
def ROL(x, n,bits):
return ROR(x, 16 - n)
what is going on and how do i fix this?
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评论（2）
好吧，看看调用
ROL(x, n, 16)时会发生什么。它调用ROR(x,16-n)，这相当于ROR(x,16-n,32)，但您真正想要的是ROR （x，16-n，16）。Well, just look at what happens when you call
ROL(x, n, 16). It callsROR(x,16-n), which is equivalent toROR(x,16-n,32), but what you really wanted wasROR(x, 16-n, 16).基本上，@GregS 的正确答案的含义是您需要在第二个实现中修复一个细节：（
我会将此作为注释，但随后我无法在其中提供可读格式的代码！-）。
Basically, the implication of @GregS's correct answers are that you need to fix one detail in your second implementation:
(I'd make this a comment, but then I couldn't have readably formatted code in it!-).