Thank you guys. On Jan 11, 9:25 pm, Thadeus Burgess <thade...@thadeusb.com> wrote: > It is because pv is a pointer to the memory that holds its list. > > When you append pv to val, your appending the *reference* to pv. > > So whenever you update pv, all references to pv also reflect these changes. > > The reason that slicing pv with [:] works is because you are making a > copy of pv, and appending the copy into val. > > The reason that pv = [i,0] works is because you assigning a new memory > reference to pv. > > -Thadeus > > On Mon, Jan 11, 2010 at 9:46 AM, DenesL <denes1...@yahoo.ca> wrote: > > e objects contain references to other objects; these are called > > containers. Examples of containers are tuples, lists and dictionaries. > > The references are part of a container's value. In most cases, when we > > talk about the value of a container, we imply the values, not the > > identities of the contained objects; however, when we talk about the > > mutability of a container, only the identities of the immediately > > contained objects are implied. So, if an immutable container (like a > > tuple) contains a reference to a mutable object, its value changes if > > that mutable object is cha
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