Dropping a unique ptr returned from a function?

Question

If I have a function source that returns a unique_ptr, and I have a function sink that calls source in the following way, it works [clang].

But is the behavior undefined? Or is everything copacetic?

class Foo {
  ...
  bool isValid() { return true; }
  ...
}

std::unique_ptr<Foo> source() {
  auto foo = std::make_unique<Foo>();  // let's pretend this is C++14?
  /* ... initialize foo */
  return foo;
}

void sink() {
  if (source()->isValid()) {
    /* do something */
  }
  /* ... */
}

Is it valid to use the unique_ptr on that line only? And when is the object theoretically supposed to be destructed? After that line? At the end of the function?

`return std::move(foo);` is redundant, you could say `return foo;`, apart from that it is valid — Piotr Skotnicki, Jul 27 '15 at 17:27
Argh, yeah. I decided partway through that a string maybe didn't make the most sense and forgot to change it. — wrhall, Jul 27 '15 at 17:40
Since it's a temporary, it's destroyed after the end of the full-expression, which in your case is the condition of the if statement. — The Paramagnetic Croissant, Jul 27 '15 at 17:41
Do you want to make that the answer? @TheParamagneticCroissant — wrhall, Jul 27 '15 at 17:41

score 4 · Answer 1 · answered Jul 27 '15 at 17:43

Yes, your code is valid. The unnamed temporary unique_ptr returned by source will be destroyed at the end of the full expression in which it was created, in this case the condition within the if statement.

From N3337, §12.2/3 [class.temporary]

... Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. ...

And from §6.4/1 [stmt.select], the grammar for an if statement is

Selection statements choose one of several flows of control.
  selection-statement:
      if ( condition ) statement
      ...
  condition:
      expression
      ...

Note that return std::move(foo); is not necessary, you can write return foo; and the unique_ptr will be automatically moved.

score 3 · Accepted Answer · answered Jul 27 '15 at 17:43

3

The result of calling a function that returns an object (and not a(n lvalue) reference) is a temporary, therefore it's alive until the end of the full expression, which is the whole condition of the if statement in this case. Hence the code is valid.

answered Jul 27 '15 at 17:43

The Paramagnetic Croissant

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Dropping a unique ptr returned from a function?

2 Answers2