Why list.sort() don't return the list reference instead of None?

[ankyhe at gmail.com]

Thanks a lot!

However, I wonder why L.sort() don't return the reference L, the
performance of return L and None may be the same.  If L.sort() return
L, we shouldn't do the awful things such as:
keys = dict.keys()
keys.sort()
for key in keys:
    ...do whatever with dict[key]...
we can only write the code as follows:
for key in dict.iterkeys().sort():
    ...do whatever with dict[key]...

Why?