Re: weird delete operator behaviour

From:

Alberto Ganesh Barbati <AlbertoBarbati@libero.it>

Newsgroups:

comp.lang.c++.moderated

Date:

Fri, 29 Jun 2007 13:17:00 CST

Message-ID:

<ibahi.51705$%k.180194@twister2.libero.it>

seaswar ha scritto:

when the destructor is commented out in the code below the output is
unexpected (an extra call to the operator delete function is being
made by the delete operator). What am I missing?? I tested on cygwin
and linux gcc version 3.4.4 with same results.

The standard is not very clear, but the latest draft in [expr.delete]
para 7 has this statement:

"If the value of the operand of the delete-expression is not a null
pointer value, the delete-expression will call a deallocation function
(3.7.3.2). Otherwise, it is unspecified whether the deallocation
function will be called."

So if the pointer is null, the deallocation function *may or may not* be
called. The fact that the destructor is trivial or not is clearly
affecting the compiler decision in you case. There is nothing wrong with
this, because the word "unspecified" doesn't require the decision to be
always the same in all cases.

HTH,

Ganesh

thanks
Suresh

output when destructor is not commented out:
$ ./a/out
~X()
operator delete()

output when destructor is commented out:
$./a.out
operator delete()
operator delete()

#include <iostream>

struct X {
    ~X()
    {
        std::cout << "~X()\n"
    }

    static void operator delete(void* p)
    {
        std::cout << "operator delete()\n";
        ::operator delete(p);
    }
};

int main()
{
    X* x = new X();
    delete x;;
    x = 0;
    delete x;
}

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