Python: A variable which is assigned and not changed changes along with another

Question

Working in Sage which is basically python, I believe. I gave the following code.

def lfsr_1(regs,tabs):
    I=regs
    leng=len(I)
    count=0
    while True:
        FB=0
        print "Before"
        print I
        print regs
        print temp
        for i in range(0,leng):
            FB=FB+tabs[i]*I[i]   //Calculating the feedback value
        for i in range(0,leng):
            regs[leng-(i+1)]=regs[leng-(i+1)-1] //Shifting regs one bit to the right
        I[0]=FB   //Adding the feedback at the end
        count=count+1  //Incrementing count value which will contain the periodicity
        print "After"
        print I
        print regs
        print temp
        if (I==regs):   //End when the initial state is repeated again. Now, count will contain the periodicity
            break

The input variables were in initialized as follows

tabs=[GF(2)(1),0,0,1,1,1]
regs=[GF(2)(0),1,1,0,1,1]
temp=regs

However, Im getting an output as:

Before
[0, 0, 1, 1, 0, 1]
[0, 0, 1, 1, 0, 1]
[0, 0, 1, 1, 0, 1]
After
[0, 0, 0, 1, 1, 0]
[0, 0, 0, 1, 1, 0]
[0, 0, 0, 1, 1, 0]

Dont know how this happens as 'I' changes along with 'regs'. 'I' is never changed in code. Is there something wrong with my assignment?

Addendum:Trying to implement a Linear Feedback Shift Register. The code is to calculate the periodicity of an LFSR. regs is the intial state, I is used to check when the regs returns to the initial state again(to calculate periodicity) and temp is just a test variable to see if another initialized variable would also be shifted.

If you would give some information on what your end goal is here, as well as comment your code, it would make it a lot easier to help you. — pzkpfw, Aug 07 '13 at 06:32

score 2 · Answer 1 · answered Aug 07 '13 at 06:36

Of course I changes along with regs because you made this assignment:

I=regs

Now I and regs reference the same list. That is how Python works. A list variable simply references a list object, it doesn't contain the whole list object.

Here's an example:

a = [1,2,3]
b = a
a[1] = 100
print b

Perhaps you wanted to make I be a copy of regs.

Try:

I = regs[:]

Of course, you might have "structural sharing" if reg contains objects.

score 1 · Answer 2 · answered Aug 07 '13 at 06:36

1

The problem lies in the line

I=regs

When you do this, you do not copy the contents of regs, you copy the reference to regs. Thus making I and regs the same array.

answered Aug 07 '13 at 06:36

Imre Kerr

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Python: A variable which is assigned and not changed changes along with another

2 Answers2