On Wednesday, 7 January 2015 at 10:56:09 UTC, John Colvin wrote:
On Wednesday, 7 January 2015 at 10:37:18 UTC, Nick wrote:
When i try to run the following code
import std.stdio;
void main(){
auto a= new test!int();
a.add(0);
a.add(1);
}
class test(T){
void add(T e){
auto temp= new node();
writeln("new node", &temp);
}
class node{
T v;
}
}
http://dpaste.dzfl.pl/c8e56b5954b8
The two nodes have the same address, is this right?
I don't know the mechanism by which they are the same
(optimiser or garbage collector), but there's not reason why
they shouldn't be.
By the time you get to the second `new node()` the first one is
completely unreachable, so why not just reuse the memory?
Considering you don't initialise them differently, the object
could even just be reused as-is.
Sorry, my mistake, bearophile is correct. Nonetheless, you
shouldn't be surprised to see memory being re-used.
