Re: Can an object know it is dead?
Noah Roberts <user@example.net> wrote:
Daniel T. wrote:
With my solution, an assert will fire if some part of the code attempts
to destroy an object that is in use. I agree though that it allows one
to attempt to use an object that was destroyed and that needs to be
fixed. Here is a revised version:
map<void*, size_t> lock;
class Object {
Object() {
assert( lock.find(this) == lock.end() );
lock[this] = 0;
}
~Object() {
assert( lock[this] == 0 );
lock.erase( this );
}
void func() {
assert( lock.find(this) != lock.end() );
++lock[this];
// do work
--lock[this];
}
};
void user( Object* o ) {
assert( lock.find(o) != lock.end() );
++lock[o];
// work with 'o'
--lock[o];
}
With the above, if you attempt to destroy an object that is still being
used by someone, an assert will fire and if you attempt to create an
object in a memory location that is in use, an assert will fire.
Of course it is wise to wrap the increment and decrement in an object
(RAII)
class Locker {
const void* obj;
Locker( const void* obj ): obj(obj) {
assert( lock.find(obj) != lock.end() );
++lock[obj];
}
~Locker() {
--lock[obj];
}
};
Your fixed version is also not portable. Any and all uses of a
deleted object illicit undefined behavior.
There is no use of any deleted object in the above code.
It also offers 0 protection in a non-debug environment. Concurrent
programs are affected by too many variables for this to work even in
cases when it might in a serial program.
All concurrent programs are by definition, non-portable so I don't see
where that has much to do with it.
What needs to be done here is to make sure the object is not deleted
before it is done being used.
Which is what my code does.
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