Re: Passing address of stack memory to placement new operator
mangesh wrote:
void someCode()
{
char memory[sizeof(Fred)];
void* p = memory;
Fred* f = new(p) Fred();
f->~Fred(); // Explicitly call the destructor for the placed
object
}
Here we r passing address of stack memory to new .
I find it baffling that you can take the time to place weird spaces
around your punctuation, but the three letters in the word "are" are
too much effort in the name of readability when you're asking others to
help you.
Anyway, yes, you're doing a placement-new on a stack-allocated array.
New is used to allocate memory on heap .
Not quite -- it allocates memory on the free store, which may or may
not be implemented in terms of the heap. It may also be overridden by
a custom allocator, which incidentally would likely make use of
placement new.
Then how can we pass address
of
stack memory to new operator . It is confusing .
Because a placement new doesn't allocate memory, it uses memory that
you've allocated for it. That's the entire point. It skips the
allocation step and goes straight to construction. It doesn't care
what part of memory the pointer points to, as long as it can be written
to.
Imagine, for example, implementing a custom allocator for small objects
based on a fixed-size arena allocated as a single chunk. That chunk
might be member data of an allocator class, and could therefore be
stack-allocated. Of course, if the allocator class was instantiated
dynamically, then its member data would live on the free store. Since
either could be the case, it would pose a real problem if the compiler
couldn't handle doing a placement new into stack-allocated memory.
Luke