What is the scope of a unique_ptr returned from a function?

Question

Would this work properly? (see example)

unique_ptr<A> source()
{
    return unique_ptr<A>(new A);
}


void doSomething(A &a)  
{  
    // ...
}  

void test()  
{  
    doSomething(*source().get());   // unsafe?
    // When does the returned unique_ptr go out of scope?
}

Answer 1

A unique_ptr returned from a function does not have scope, because scope only applies to names.

In your example, the lifetime of the temporary unique_ptr ends at the semicolon. (So yes, it would work properly.) In general, a temporary object is destroyed when the full-expression that lexically contains the rvalue whose evaluation created that temporary object is completely evaluated.

Answer 2

Temporary values get destroyed after evaluating the "full expression", which is (roughly) the largest enclosing expression - or in this case the whole statement. So it's safe; the unique_ptr is destroyed after doSomething returns.

Answer 3

It should be fine. Consider

int Func()
{
  int ret = 5;

  return ret;
}

void doSomething(int a) { ... }


doSomething(Func());

Even though you're returning ret on the stack it is okay because it's within the calling scope.