context not cleaned at the end of a loop containing yield?

[TP]

Hi everybody,

This example gives strange results:

########
def foo( l = [] ):

l2 = l
print l2
for i in range(10):
if i%2 == 0:
l2.append( i )
yield i
########

[i for i in ut.foo()]

[]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8, 0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
....

How to explain this behavior? Why l is not [] when we enter again the
function foo()?

Thanks

Julien

--
python -c "print ''.join([chr(154 - ord(c)) for c in '*9(9&(18%.\
9&1+,\'Z4(55l4('])"

"When a distinguished but elderly scientist states that something is
possible, he is almost certainly right. When he states that something is
impossible, he is very probably wrong." (first law of AC Clarke)

 

[MRAB]

Hi everybody,

This example gives strange results:

########
def foo( l = [] ):

l2 = l
print l2
for i in range(10):
if i%2 == 0:
l2.append( i )
yield i
########

[i for i in ut.foo()]

[]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8, 0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
...

How to explain this behavior? Why l is not [] when we enter again the
function foo()?

Read http://www.ferg.org/projects/python_gotchas.html#contents_item_6

 

[Mensanator]

Hi everybody,

This example gives strange results:

########
def foo( l = [] ):

    l2 = l
    print l2
    for i in range(10):
        if i%2 == 0:
            l2.append( i )
        yield i
########

[i for i in ut.foo()]

[]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in ut.foo()]

[0, 2, 4, 6, 8, 0, 2, 4, 6, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
...

How to explain this behavior? Why l is not [] when we enter again the
function foo()?

Readhttp://www.ferg.org/projects/python_gotchas.html#contents_item_6- Hide quoted text -

So, l=[] only happens once, not everytime you call it.

Also, 12 = l means 12 is the same object and remembers it's contents
across calls
(since l is never reset).

Try this:

def foo(l=[0]):
l2 = l[1:] # deep copy, omitting call count
print l2,l2 is l,l
l[0] += 1 # count how many times we call foo
for i in range(10):
if i%2 == 0:
l2.append(i)
yield i

[i for i in foo()]

[] False [0]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in foo()]

[] False [1]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[i for i in foo()]

[] False [2]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

You can see that l is not being reset to [0] on subsequent calls
and that 12 is a seperate list from l, so appending to l2 does not
change l.