C# out-parameter

What is the difference between the following two code snippets?

public void foo(out classA x)
{
    y = new classA();
    x = y;
}

and the second:

public void foo(out classA x)
{
    classA y;
    x = y;
}

Is it true that the second snippet is dangerous, because x has now a reference to this local y, which might be already dea开发者_运维知识库d after exiting foo?

Why do we have to use "new" in general?

I'm a bit confused, because in C++, if x in the second snippet would be a pointer, the statement x = y would not even compile since y is not a pointer.

---

Let's assume for a moment that the second snippet is

public void foo(out classA x)
{
  classA y = new classA();
  x = y;
}

The snippet as you wrote it does not compile in C# at all. y has to be assigned first. The line classA y; does not create an instance of classA on the stack as it would in C++. It simply declares a variable y of type classA.

With the compiling snippet in mind, assigning an out variable to a locally declared and initialized object is not dangerous in any way. The classA object that y points to and that is assigned to x will remain alive and well until x goes out of scope wherever it's declared/used.

---

Your confusion, I think, is that in C++ the second example would return a reference to a stack-allocated object. That doesn't happen in C#.

Given your second example in C#:

public void foo(out classA x)
{
  classA y;  // Doesn't allocate an object
  x = y;
}

This probably won't compile, anyway, because y is never assigned a value.

---

Is it true that the second snippet is dangerous [...]

Not in C#; you're thinking of native languages like C or C++ that allow one to take a pointer to a local variable. In C#, everything (within reason) is a reference to a managed object. Everything is essentially on the heap.

---

It is not dangerous because the code will NOT be able to be compiled unless all "out" parameters are given a definition before the method returns or exits.