Hi Jonathan & Everyone,
Thanks for your reply. Yes, as you pointed out, it
will create only 2 temporaries as below:
X a(1);
sub r3, fp, #16
mov r0, r3
mov r1, #1
bl X::X[in-charge](int)
a = f(a);
sub r3, fp, #24
sub r2, fp, #16
mov r0, r3
mov r1, r2
bl X::X[in-charge](X const&)
sub r2, fp, #24
sub r3, fp, #20
mov r0, r3
mov r1, r2
bl f(X)
sub r3, fp, #20
mov r0, r3
bl X::~X [in-charge]()=>dtor for temp1
sub r3, fp, #24
mov r0, r3
bl X::~X [in-charge]()=>dtor for temp2
sub r3, fp, #16
mov r0, r3
bl X::~X [in-charge]()=>dtor for a
What i want to know is how to find what are the
temporaries inserted by the compl.
I tried looking into the ABI but couldn't get much
help. Is there any way i can figure out the
temporaries inserted by the compl.
Looking forward for more help. :)
Best regards,
San
--- Jonathan Wakely <[EMAIL PROTECTED]> wrote:
> On Thu, Feb 10, 2005 at 01:49:44PM +0000, Jonathan
> Wakely wrote:
>
> > > a = f(a); //this will create 4 temorary objects
> >
> > Why? I think this will only create two
> temporaries.
> >
> > One when a is copied to the "x" parameter of f()
> and one when "x" is
> > copied to the return value;
>
> to be clear, "x" isn't a true temporary as it has a
> name, but it is
> created and destroyed during the execution of f(),
> which is what I
> assume you meant.
>
> jon
>
> --
> "Keep away from people who try to belittle your
> ambitions. Small people
> always do that, but the really great make you feel
> that you, too, can
> become great."
> - Mark Twain
>
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