Are here any benefits from using std::unique_ptr?

user_10171966 0
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I stumbled upon code like this:

void foo(T* bar); // Note: takes ownership of 'bar'.

foo(new T());

And now I wonder if there is any point in refactoring it to:

void foo(T* bar); // Note: takes ownership of 'bar'.

auto tempT = std::make_unique<T>();
foo(tempT.release());
  • Is it more exception-safe?
  • It certainly adds a little bit more clarity regarding the transfer of ownership, though calling 'new' from the argument list already makes that very clear by itself.

Note that I unfortunately can't change the signature of 'foo'.

unique-ptr exception c++
3 Answers

Is it more exception-safe?

No. However, consider a slightly more complex example:

auto tempT = std::make_unique<T>();
some_operation();
foo(tempT.release());

In this case, there would be a potential issue with exception safety if unique_ptr wasn't used.

That said, much safer would be:

void foo(std::unique_ptr<T> bar);  // no need to note that takes ownership
                                   // because that's implied by the type

Note that I unfortunately can't change the signature of 'foo'.

Then write a wrapper function that you can control.

user_20793030

No it is not more exception safe.

But consider if function changes to take more parameters:

void foo(T* bar, T* more_bar = get_more_bar()); // Note: takes ownership of both.

foo(new T());

Now what if get_more_bar() is evaluated after new T() and throws.

user_29450270

You might consider writing a wrapper for foo:

void wrap_foo(std::unique_ptr<T> arg) {
    foo(arg.get());
    arg.release();
}

If foo were to throw an exception without freeing the memory, this would free it. However, if foo is already exception-safe and frees the memory before it throws an exception, this would free it twice. Unfortuately there's no good way to write a safe wrapper for this that will detect if the memory has been released properly or not.

user_164060
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