Python Simple Swap Function

As it is, your final print will print out the original values of s1 and s2. This is because you're only swapping them within the scope of the function. Doing so will not affect their values outside the function (i.e. after their values after the function has been called)

If they are mutable types (list, set, dict, etc), then you could modify them in-place inside swap. However, that restricts swap to work only on mutable types.

You are therefore better off returning the inputs in reversed order:

def swap(s1, s2):
    return s2, s1

s1 = 'a'
s2 = 'b'
s1, s2 = swap(s1, s2)
print s1, s2 # prints 'b a'

Of course, you could do this all in one line as follows:

s1, s2 = s2, s1

Cheers!

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.