auto_ptr safety

[Raider]

Is it exception-safe to write like this:

void foo(auto_ptr<Tx )
{
...
}

void bar()
{
foo(auto_ptr<T> (new T(...));
}
Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

[red floyd]

Raider wrote:

Is it exception-safe to write like this:

void foo(auto_ptr<Tx )
{
...
}

void bar()
{
foo(auto_ptr<T> (new T(...));
}
Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

It depends. You do know that anything you pass to foo loses ownership
of the pointer, is that what you want? Or did you want to pass an
auto_ptr<T>&?

Given the ownership transfer, I'd use the first example of bar.

[Alan Johnson]

Raider wrote:

Is it exception-safe to write like this:

void foo(auto_ptr<Tx )
{
...
}

void bar()
{
foo(auto_ptr<T> (new T(...));
}
Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

In your specific example the usage is exception safe (so far as I can
tell). However, it isn't a bad habit to get into to initialize
auto_ptrs (and shared_ptrs) in a separate variable. Consider what
could happen if you had another parameter.

void foo(auto_ptr<Tx , int y)
{
}

And now you call it using the same style:

int f() { /* whatever */ }

void bar()
{
foo(auto_ptr<T> (new T), f()) ;
}
You've now lost exception safety, depending on how the compiler decides
to order the various operations in the expression. Consider this
ordering:
(1) new T
(2) f()
(3) auto_ptr<T>()
(4) foo()

If (1) succeeds, but (2) throws an exception, then you leak memory.
This could not happen if you used a named auto_ptr instead of a
temporary. If you really hate having your auto_ptrs be automatic
variables, one idiom is to write a version of new that returns an
auto_ptr. Example:

template <typename T>
auto_ptr<Tauto_ new()
{
return auto_ptr<T>(new T) ;
}

Then you could call:

foo(auto_new<T> (), f()) ;

The difference between this and the previous attempt is that, while the
compiler may reorder the parts of an operation, it will not reorder
parts of a function (the 'as if' clause notwithstanding ), so you are
guarantee that if new succeeds, the result gets put into an auto_ptr
before anything else happens.

The downside of the above approach is that it is not as flexible as
new. Notably, it would only work with constructors that take no
parameters. You could create an overload or separate function for each
type of constructor, but whether that is more or less work than just
making auto_ptr's named variables depends on the details of your
situation.

--
Alan Johnson

[Michael]

Is it exception-safe to write like this:

>
void foo(auto_ptr<Tx );

void bar()
{
foo(auto_ptr<T> (new T(...));
}

Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

I'd go with the second. Right now it's a toss up, but a month from
now, when someone redefines foo to take 2 pointers:
void foo(auto_ptr<Tx , auto_ptr<Ty);
then you're going to want the second. So I'd code up the second way
now, to make my code better in the sense that it's less likely to be
broken if someone modifies it later.

See http://www.gotw.ca/gotw/056.htm for a more detailed discussion.

Michael

[Raider]

Alan Johnson wrote:

Raider wrote:

>Is it exception-safe to write like this:

void foo(auto_ptr<Tx )
{
...
}

void bar()
{
foo(auto_ptr<T> (new T(...));
}
Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

In your specific example the usage is exception safe (so far as I can
tell). However, it isn't a bad habit to get into to initialize
auto_ptrs (and shared_ptrs) in a separate variable. Consider what
could happen if you had another parameter.

Thank you Alan! It's exactly what I asked for. I remember that its bad
habbit to pass `new T(...)' to the function that takes auto_ptr<T>, but
I've forgot why. Thanks again.

[bb]

Hi Alan,
Beautiful explanation. Thanks. I have another question.

Is the following safe? (it works fine though).

void thirdPartyLibFu n(T* t) {
...
}

void myFun() {
std::auto_ptr<T ap2t = std::auto_ptr<T >(new T);

thirdPartyLibFu n(ap2t.get());
}

Cheers.
Alan Johnson wrote:

Raider wrote:

Is it exception-safe to write like this:

void foo(auto_ptr<Tx )
{
...
}

void bar()
{
foo(auto_ptr<T> (new T(...));
}
Or should I create temporary variable?

void bar()
{
auto_ptr<Tx(new T(...);
foo(x);
}

In your specific example the usage is exception safe (so far as I can
tell). However, it isn't a bad habit to get into to initialize
auto_ptrs (and shared_ptrs) in a separate variable. Consider what
could happen if you had another parameter.

void foo(auto_ptr<Tx , int y)
{
}

And now you call it using the same style:

int f() { /* whatever */ }

void bar()
{
foo(auto_ptr<T> (new T), f()) ;
}
You've now lost exception safety, depending on how the compiler decides
to order the various operations in the expression. Consider this
ordering:
(1) new T
(2) f()
(3) auto_ptr<T>()
(4) foo()

If (1) succeeds, but (2) throws an exception, then you leak memory.
This could not happen if you used a named auto_ptr instead of a
temporary. If you really hate having your auto_ptrs be automatic
variables, one idiom is to write a version of new that returns an
auto_ptr. Example:

template <typename T>
auto_ptr<Tauto_ new()
{
return auto_ptr<T>(new T) ;
}

Then you could call:

foo(auto_new<T> (), f()) ;

The difference between this and the previous attempt is that, while the
compiler may reorder the parts of an operation, it will not reorder
parts of a function (the 'as if' clause notwithstanding ), so you are
guarantee that if new succeeds, the result gets put into an auto_ptr
before anything else happens.

The downside of the above approach is that it is not as flexible as
new. Notably, it would only work with constructors that take no
parameters. You could create an overload or separate function for each
type of constructor, but whether that is more or less work than just
making auto_ptr's named variables depends on the details of your
situation.

--
Alan Johnson