python中位的循环移位（相当于Fortran的ISHFTC）

Question

I want to achieve the equivalent of the ISHFTC function in Fortran using python. What is the best way to do this?

For example,

x = '0100110'
s = int(x, 2)
s_shifted = ISHFTC(s,1,7) #shifts to left by 1
#binary representation of s_shifted should be 1001100

My attempt based on Circular shift in c

def ISHFTC(n, d,N):  
    return (n << d)|(n >> (N - d)) 

However, this does not do what I want. Example,

ISHFTC(57,1,6) #57 is '111001'

gives 115 which is '1110011' whereas I want '110011'

Answer 1

Your attempted solution does not work because Python has unlimited size integers.

It works in C (for specific values of N , depending on the type used, typically something like 8 or 32), because the bits that are shifted out to the left are automatically truncated.

You need to do this explicitly in Python to get the same behaviour. Truncating a value to the lowest N bits can be done be using % (1 << N) (the remainder of dividing by 2 ^N ).

Example: ISHFTC(57, 1, 6)

We want to keep the 6 bits inside |......| and truncate all bits to the left. The bits to the right are truncated automatically, because the these are already the 6 least significant bits.

n                  |111001|
a = n << d        1|110010|
m = (1 << N)      1|000000|
b = a % m         0|110010|

c = n >> (N - d)   |000001|(11001)

result = b | c     |110011|

Resulting code:

def ISHFTC(n, d, N):  
    return ((n << d) % (1 << N)) | (n >> (N - d))
          #  ^^^^^^ a
          #             ^^^^^^ m
          #  ^^^^^^^^^^^^^^^^^ b
          #                         ^^^^^^^^^^^^ c

>>> ISHFTC(57, 1, 6)
51
>>> bin(_)
'0b110011'