Are ref and out in C# the same a pointers in C++?

Question

I just made a Swap routine in C# like this:

static void Swap(ref int x, ref int y)
{
    int temp = x;
    x = y;
    y = temp;
}

It does t

Answer 1

The nice thing about using out is that you're guaranteed that the item will be assigned a value -- you will get a compile error if not.

Answer 2

Actually, I'd compare them to C++ references rather than pointers. Pointers, in C++ and C, are a more general concept, and references will do what you want.

All of these are undoubtedly pointers under the covers, of course.

Answer 3

While comparisons are in the eye of the beholder...I say no. 'ref' changes the calling convention but not the type of the parameters. In your C++ example, d1 and d2 are of type int*. In C# they are still Int32's, they just happen to be passed by reference instead of by value.

By the way, your C++ code doesn't really swap its inputs in the traditional sense. Generalizing it like so:

template<typename T>
void swap(T *d1, T *d2)
{
    T temp = *d1;
    *d1 = *d2;
    *d2 = temp;
}

...won't work unless all types T have copy constructors, and even then will be much more inefficient than swapping pointers.

Answer 4

The short answer is Yes (similar functionality, but not exactly the same mechanism). As a side note, if you use FxCop to analyse your code, using out and ref will result in a "Microsoft.Design" error of "CA1045:DoNotPassTypesByReference."

Answer 5

Reference parameters in C# can be used to replace one use of pointers, yes. But not all.

Another common use for pointers is as a means for iterating over an array. Out/ref parameters can not do that, so no, they are not "the same as pointers".

Answer 6

ref and out are only used with function arguments to signify that the argument is to be passed by reference instead of value. In this sense, yes, they are somewhat like pointers in C++ (more like references actually). Read more about it in this article.