id(a) == id(b) and a is not b --> bug?

[Martin v. Löwis]

Gerrit Holl <gerrit at nl.linux.org> writes:

> This would mean that "id(a) == id(b)" is the same as "a is b".

It doesn't, atleast not if a and b can be arbitrary expressions:

count=0
def foo():
  global count
  count+=1
  if count<4:
    return 1
  return 2

print id(foo())==id(foo())
print foo() is foo()

If any kind of computation is performed in the arguments, the property
you assume may not hold.

Regards,
Martin