Question

I came across this problem when attempting to learn python. Consider the following function:

def swap0(s1, s2):
    assert type(s1) == list and type(s2) == list
    tmp = s1[:]
    s1 = s2[:]
    s2 = tmp
return

s1 = [1]
s2 = [2]
swap0(s1, s2)
print s1, s2

What will s1 and s2 print?

After running the problem, I found that the print statement will print 1 2. It seems that the value of s1 and s2 did not change from the swap0 function. The only explanation that I could think of was because of the line.

tmp = s1[:]

Since s1[:] is a copy, this makes sense that the value of s1 will not change in the function call. However because the parameter of swap0 is (s1, s2), I am not sure if after doing tmp = s1[:]. Anytime I do

s1 = something...

it will be a reference to the copy of s1, instead of s1 itself. Can someone offer a better explanation? Thanks.

Answer 1

It’s because it assigns new values to s1 and s2 inside the swap0 function. These assignments do not propagate outside the function. You’ll see that it works if you just copy and paste the function body in the place of the function call.

You can work around this by modifying the objects referenced by the arguments, rather than the arguments themselves:

def swap0(s1, s2):
    assert type(s1) == list and type(s2) == list
    tmp = s1[:]
    s1[:] = s2
    s2[:] = tmp

However, the easier and better way to do a swap in Python is simply:

s1, s2 = s2, s1

This, too, will only swap those particular references to the lists, but not the list contents themselves.