Shane Hathaway wrote:
> Michael wrote:
>> sorry, I'm used to working in c++ :-p
>>
>> if i do
>> a=2
>> b=a
>> b=0
>> then a is still 2!?
>>
>> so when do = mean a reference to the same object and when does it
>> mean make a copy of the object??
>
> To understand this in C++ terms, you have to treat everything,
> including simple integers, as a class instance, and every variable is
> a reference (or "smart pointer".) The literals '0' and '2' produce
> integer class instances rather than primitive integers. Here's a
> reasonable C++ translation of that code, omitting destruction issues:
>
> class Integer
> {
> private:
> int value;
> public:
> Integer(int v) { value = v; }
> int asInt() { return value; }
> }
>
> void test()
> {
> Integer *a, *b;
> a = new Integer(2);
> b = a;
> b = new Integer(0);
> }
>
A closer translation would be:
const Integer CONST0(0);
const Integer CONST2(2);
void test()
{
const Integer *a, *b;
a = &CONST0;
b = a;
b = &CONST2;
}
The constant integers are created in advance, not when you do the
assignment. Arithmetic may create new Integer objects, but when the result
is a small integer it simply reuses an existing object.
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