------- Comment #6 from fang at csl dot cornell dot edu 2007-03-08 22:58
-------
Subject: Re: Overloaded operator delete[] doesn't get called
This following test case is 'interesting':
---------------->8 snip 8<---------------------
#include <iostream>
using std::cout;
class one_array_only {
private:
static one_array_only pool[256];
public:
static void * operator new[] (std::size_t size) throw() {
cout << "class new[] operator\n";
#if WTF
return pool;
#else
return NULL;
#endif
}
static void operator delete[] (void *p) throw() {
cout << "class delete[] operator\n";
}
};
one_array_only
one_array_only::pool[256];
int
main(int argc, char* argv[]) {
one_array_only* p = new one_array_only[12];
delete [] p;
return 0;
}
---------------->8 snip 8<---------------------
Above, in operator new[], If WTF is false, returning NULL, I reproduce the
same error (missing call to class operator delete []). If WTF is true
(returning pool), the right operator delete [] is called.
// output with WTF false
class new[] operator
// output with WTF true
class new[] operator
class delete[] operator
Tested on: powerpc-apple-darwin8-g++-4.0.1
Why on earth would the definition inside operator new[] influence whether
or not operator delete[] is called?
Fang
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=29164
