digitalmars.D - new or no?
- Shadow_exe (24/24) Dec 23 2012 for(uint y=0; y<10; ++y){
- Adam D. Ruppe (10/14) Dec 23 2012 That's the address of the local variable. The object it points to
- Andrej Mitrovic (4/8) Dec 23 2012 Use writeln(cast(void*)m) to get the address of the object, otherwise
- Shadow_exe (2/2) Dec 23 2012 Yes, thank you!
for(uint y=0; y<10; ++y){
auto m = new Mutex();
writeln(&m);
}
run:
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
7F66E5A05CF8
As I understand it, all the time return one object
But:
auto m = new Mutex();
writeln(&m);
auto m1 = new Mutex();
writeln(&m1);
run:
7FCF60698CF0
7FCF60698CF8
Dec 23 2012
On Sunday, 23 December 2012 at 19:43:54 UTC, Shadow_exe wrote:
for(uint y=0; y<10; ++y){
auto m = new Mutex();
writeln(&m);
}
That's the address of the local variable. The object it points to
is somewhere else.
An Object in D is more like an Object* in C++.
Object* o = new Object();
&o == 0
o == 1
o = new Object();
&o == 0 // the local variable is still in the same place
o == 2 // but it now points to a new object
Dec 23 2012
On 12/23/12, Shadow_exe <shadow_exe ukr.net> wrote:
for(uint y=0; y<10; ++y){
auto m = new Mutex();
writeln(&m);
}
Use writeln(cast(void*)m) to get the address of the object, otherwise
you're writing the address of the reference (which each time refers to
a different object).
Dec 23 2012
Yes, thank you! Tired apparently, I need to rest...
Dec 23 2012









"Adam D. Ruppe" <destructionator gmail.com> 