Re: diff between user defined delete and delete[]
On Nov 17, 10:13 am, mail....@gmail.com wrote:
Suppose in a class we overload four operators:
operator new
operator delete
operator new[]
operator delete[]
class Test{
public:
void * operator new (size_t t){
cout<<"\nCalling... new";
return malloc(t);
}
void operator delete (void *p){
cout<<"\nCalling ... delete";
free(p);
}
void * operator new [] (size_t t){
cout<<"\nCalling ... new[]";
return malloc(t);
}
void operator delete [] (void *p){
cout<<"\nCalling ... delete[]";
free(p);
}
};
If we perform
Test *p=0; delete p;
It calls operator delete.
But if we perform
Test *p=0; delete []p;
It doesn't call operator delete[] until and unless we don't
call operator new[].
I'm not sure how to read the above. Too many negations. But in
a delete expression, if the pointer is null, it is unspecified
whether the compiler calls the operator delete function or not.
Probably, your compiler calls it in the simple cases, but
doesn't bother calling it if it otherwise has to check for a
null pointer (usually the case with delete[]).
Means If we do like this;
Test *p=0; p=new Test[10]; delete[] p; It calls operator delete[]=
..
Why???
Why what? If the pointer isn't null, the compiler is required
to call operator delete. If the pointer is null, it is
unspecified whether operator delete is called or not (so the
operator delete function had better check). And it can
sometimes do one, sometimes the other---adding a virtual
destructor (or even any non-trivial destructor) might change
the behavior, for example.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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