So immutable objects cannot be modified directly? I guess this means integers are immutable and the act of assigning to one is a completely new definition? So if I were to create a class called Integer and give it a .set() method, this would allow me to create mutable integers, and thus passing in an object of type class Integer would allow me to modify the value from inside the function?
On 8/16/07, Steve Holden <[EMAIL PROTECTED]> wrote: > > Robert Dailey wrote: > > Hi, > > > > I previously created a topic named "Pass by reference or by value" where > > I inquired on how python's function parameters work. I received a lot of > > nice responses, however I'm still confused on the topic. Note that I > > come from a C++ background to Python, so any comparisons to C++ would be > > very helpful. > > > > I ran a few tests. There's two tests in particular I wanted to show you > > guys: > > > ------------------------------------------------------------------------------------------------ > > myvar = [] > > > > def changeme( param ): > > param.append( "blah" ) > > print param > > > > changeme( myvar ) > > > > print myvar > > > > The above code yields the following output: > > ['blah'] > > ['blah'] > > > > This means that the list passed in was modified by the function. > > > ------------------------------------------------------------------------------------------------ > > Now test case 2: > > > > myvar = 4 > > > > def changeme( param ): > > param = 5 > > print param > > > > changeme( myvar ) > > > > print myvar > > > > The above code yields the following output: > > 5 > > 4 > > > > This means that the integer passed in was NOT modified by the function. > > > ------------------------------------------------------------------------------------------------ > > > > Between these two tests, both types passed in are mutable objects. I'm > > having trouble figuring out what mandates an object to be changed from > > within a function versus not. What is happening in test case 2 to cause > > it to not be modified? > > > > Thanks for reading guys. Hopefully one day I'll understand this lol. > > > The first thin to realise is that all Python names are just bindings to > values. In C++ terms you can think of them all as pointers. > De-referencing is automatic when a value is to be retrieved. > > > >>> def changeme( param ): > ... param = 3 > ... print param > ... > >>> myvar = [] > >>> changeme(myvar) > 3 > >>> myvar > [] > >>> > > In this case there is no attempt to mutate the argument, the argument > name is simply bound to another value. Since the argument is a name > local to the function, this does not result in any change outside the > function. > > In this case the argument is bound to a mutable value, so the call to > append mutates the object (a list) referenced by the argument. > > Unlike C++ and similar languages a variable does not hold a value, it > holds a pointer to a value. When a list is passed as a function argument > the reference to the list is copied into the argument. Does this help at > all? > > regards > Steve > -- > Steve Holden +1 571 484 6266 +1 800 494 3119 > Holden Web LLC/Ltd http://www.holdenweb.com > Skype: holdenweb http://del.icio.us/steve.holden > --------------- Asciimercial ------------------ > Get on the web: Blog, lens and tag the Internet > Many services currently offer free registration > ----------- Thank You for Reading ------------- > > -- > http://mail.python.org/mailman/listinfo/python-list >
-- http://mail.python.org/mailman/listinfo/python-list
