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.