将唯一指针的指针传递给采用双指针的 function

Question

I have a function that effectively does this

void foo(Class** c)
{
    // memory checks and stuff
    (*c) = new Class();
    // more stuff
}

I cannot change this function. To call this function I have to do something like this.

Class* c = nullptr;
foo(&c);
if (c)
{
    // stuff
}
delete c;

I would very much prefer to use a std::unique_ptr rather than the raw pointer. However, I don't know how to get the address of the internal pointer. The listing below does not compile, obviously, because I'm trying to take the address of an rvalue.

std::unique_ptr<Class> c = nullptr;
foo(&(c.get()));
if (c)
{
    // stuff
}

I realize I could make the raw pointer as well as the unique pointer, call the function, then give the raw pointer to the unique pointer, but I would prefer to not. Is there a way to do what I want to do?

Answer 1

Create a wrapper around foo :

std::unique_ptr<Class> foo()
{
    Class* c = nullptr;
    foo(&c);
    return std::unique_ptr<Class>(c);
}

Answer 2

Your hands are tied by the API of the function.

The best solution I personally see is to do what you said you'd rather not: create the unique_ptr after calling the function.

If you call this function a lot or if you have many functions like it I would create a wrapper function that creates locally the raw pointer and returns unique_ptr.