Using unsafe will be faster?

[Abubakar]

Hi,
NOTE: in the following code Rectangle is a struct inside the System.Drawing
namespace.

I'v a following code:
Rectangle GetFillRectangle(ref Rectangle rect)
{

Rectangle newrect;
//calculate fill dimension for the rect passed and return the new one

return newrect;
}

If I write it using pointers will it be faster? Like:
unsafe Rectangle * GetFillRectangle(ref Rectangle rect)
{
Rectangle newrect;
//calculate fill dimension for the rect passed and return the new one

return & newrect;

}

My confusion: when I write the following code (using the first version of
the function):

....
Rectangle fillrect= GetFillRectangle(ref somerectangle);

fillrect, above, will get a "copy" of the newrect or just a reference of the
newrect created inside the GetFillRectangle() function? My understanding
says that it'll get a "copy" of the newrect because its a value type, and if
this is so there are actually 2 rectangle objects being created and only one
is getting used the first one gets gc. And if this all is true than if I
write the following code:
Rectangle * rect = GetFillRectangle(ref somerectangle);

rect pointer will get a direct reference to the newrect rectangle object
created inside the GetFillRectangle() function. So I'm saved from extra time
& memory that would have taken to create a copy of the rectangle object. Is
that true?

[Blomberg]

What you say is alright. But instead of using unsafe code, you can achieve same effect by using an out parameter.

[Abubakar]

oh thats greate, thanx I just forgot out, I'll use that.

"Blomberg" <an*******@discussions.microsoft.com> wrote in message
news:EF**********************************@microsof t.com...

What you say is alright. But instead of using unsafe code, you can achieve

same effect by using an out parameter.