Re: Is new observable behavior?

From:
"Balog Pal" <pasa@lib.hu>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 2 Apr 2009 14:02:35 CST
Message-ID:
<gr2ir1$20hb$1@news.ett.com.ua>

"Greg Herlihy" <greghe@mac.com>

On Apr 1, 2:54 pm, Chris <cuz...@gmail.com> wrote:
void * operator new(std::size_t size) {
   std::cerr << "Observable? (new)" << std::endl;
   return std::malloc(size);

}

void operator delete(void * ptr) {
   std::cerr << "Observable? (delete)" << std::endl;
   std::free(ptr);
}
...
Here's the output:

./a.out
Observable? (new)
Observable? (delete)
test_new_optimization(3) is `3'

Thus, it's observable.


Thus, it was MADE observable, by inserting library I/O functions.

These two particular implementations of new() and delete() have
observable behavior. In general, though, new() and delete() do not
perform any I/O or have any other kind of observable behavior. So the
answer to the original question, is - for all practical purposes -
"yes" - the compiler will almost certainly be free to eliminate the
calls to new() and delete().


Especially they have fully defined semantics, so the compiler knows
everything to do the job.

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