Re: Why does "delete" not delete?
On Mar 10, 8:40 pm, Lance Diduck <lancedid...@nyc.rr.com> wrote:
On Mar 10, 12:27 pm, "D. Susman" <derya.sus...@gmail.com> wrote:
On Mar 10, 6:20 pm, alasham.s...@gmail.com wrote:
On Mar 10, 5:11 pm, "D. Susman" <derya.sus...@gmail.com> wrote:
When I call "clean" as shown below, shouldn't I be
getting segmentation fault due to garbage data? I am
compiling with CC and the output is 5. Does "delete" not
delete in this situation or is it no guarantee that it
will print 5 or go wrong?
void clean( int* ptr )
{
delete ptr;
}
//...
int* ptr = new int( 5 );
clean( ptr );
cout << *ptr << endl;
It is 'undefined behaviour', so the result may vary from
one implementation to another, and on different runs of
the same program.
Thank you. Is there a way to make a pointer undeletable when
passed to a function?- Hide quoted text -
There are a number of ways. This is what I do:
void clean( int& ptr )
{
// delete ptr; //wont compile
But "delete &ptr" will.
ptr=0;
}
std::auto_ptr<int> ptr(new int(5));
clean(*ptr);
cout<<*ptr<<endl;
prints '0' always
In the end, a function must fulfill a defined contract. If you
don't know what the contract is, you don't use the function. If
the function's contract says it deletes the memory, you only
pass it pointers to stuff you want deleted. And if the
function's contract doesn't say that it deletes the memory, then
it had better not delete it, since you're quite likely to pass
it the address of a local variable, or whatever.
There are only a very limited number of contractual constraints
that can be expressed in the language. As a programmer, you
need to respect all of the contractual constraints, not just
those expressed in the language.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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