On Fri, 20 Jul 2012 14:50:19 +0100, Namespace <rswhi...@googlemail.com>
wrote:
New question:
I have this code:
[code]
import std.stdio;
struct Test {
public:
this(int i = 0) {
writeln("CTOR");
}
this(this) {
writeln("COPY CTOR");
}
~this() {
writeln("DTOR");
}
}
void main() {
Test[] _arr;
_arr ~= Test(0);
writeln("end main");
}
[/code]
And as output i see:
CTOR
COPY CTOR
DTOR
end main
Why on earth....?
I create a struct Test. It's not a local variable, it's directly
assigned,
but it is copied and the original is destroyed. Why?
Because that's how assignment works, rhs is evaluated, struct is
constructed, then assigned to lhs, lhs copies the variable into the
array. Your variable is a struct, so it is copied. Use a pointer instead
and the pointer is copied into the array. Use a reference and the
reference is copied into the array. The variable, whatever it is, is
/always/ copied into the array.
It's a long standing optimisation issue and the reason for the recent
distinction between lvalues and rvalues made in C++0x for their new 'move'
optimisation where assignments involving rvalues/temporaries can move the
rvalue instead of copying - or something like that, I'm a bit vague having
watched a vid a few days back and not paying a lot of attention to it,
sorry.
Does:
_arr[0] = Test(0);
avoid the copy construction?
R
