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From: clarkcox3@gmail.com ("Clark S. Cox III")
Newsgroups: comp.std.c++
Subject: Re: auto_ptr as a return value
Date: Wed, 24 Aug 2005 19:52:09 GMT
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On 2005-08-24 06:41:12 -0400, andru123@hotmail.com said:

> auto_ptr<Bla> makeBla()
> {
>  return NULL;
> }

First, auto_ptr's constructor is explicit, which means that NULL will 
not be implicitly converted in this case (I'm assuming that in your 
real code, you're returning an actual object.

> //Assuming auto_ptr is OWNER of the objects, I MUST call release()
> auto_ptr<Bla> bla = makeBla().release();

You don't have to do that at all. The assignment operator and copy 
constructor of auto_ptr both transfer the ownership of the contained 
pointer:

[littleclark2:~] clarkcox% cat test.cpp
#include <memory>
#include <iostream>

using namespace std;

struct Foo
{
  Foo() { std::cout << "Foo::Foo() called.\n"; }
  ~Foo() { std::cout << "Foo::~Foo() called.\n"; }
};

auto_ptr<Foo> makeBla()
{
  return auto_ptr<Foo>(new Foo);
}

int main()
{
  auto_ptr<Foo> a = makeBla();
  Foo           *raw = a.get();
  assert(a.get() == raw);

  auto_ptr<Foo> b = a;
  assert(a.get() == NULL);
  assert(b.get() == raw);


  auto_ptr<Foo> c;
  c = b;
  assert(a.get() == NULL);
  assert(b.get() == NULL);
  assert(c.get() == raw);

  std::cout  << "a, b and c are about to go out of scope.\n";
  return 0;
}

[littleclark2:~] clarkcox% c++ test.cpp && ./a.out
Foo::Foo() called.
a, b and c are about to go out of scope.
Foo::~Foo() called.

[littleclark2:~] clarkcox%


> 
> 
> As you can guess, here i get an exception.
> 
> Can I do this then:
> 
> auto_ptr<Bla> bla = NULL;
> if (makeBla() != NULL)
> {
>  bla = makeBla();
> }
> 
> My question:
> will the result of the first call of makeBla() be deleted/ handled
> correctly?

Yes, it will be deleted when the unnamed, temporary auto_ptr is 
destructed, however, as I explained above, your problem is elsewhere, 
as auto_ptr is explicitly designed to handle this case.


-- 
Clark S. Cox, III
clarkcox3@gmail.com

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