C#

Ennis Ray Lynch, Jr. wrote:

I think you suffer from a fundamental misunderstanding of pointers and dynamic memory. Spend some time in c and come back to the subject when you are refreshed.

Ennis Ray Lynch, Jr. wrote:

There is a difference between passing a reference and passing a reference to a reference and I would have thought someone with assembly language experience would have known that.

struct S
{
   void SayHello() {}
}

void Caller()
{
   S *s = new S();
   Callee(s);
}

void Callee(S *p)
{
   p->SayHello();
}

S. Senthil Kumar wrote:

Maybe you're assuming references to behave like C++ references?

S. Senthil Kumar wrote:

Will you agree that the instance pointed at by s is passed by reference?

Guffa wrote:

No, it's not a copy of the instance that is passed. It's the value of the variable s that is passed, which happens to be a pointer to the instance.

Guffa wrote:

Was that code supposed to show an example of a C++ reference? Because it isn't.

S. Senthil Kumar wrote:

Well, than can you write a snippet of code that demonstrates passing by reference?

void GetAnswer(ref string answer) {
   answer = "42";
}

string answer = null;
GetAnswer(ref answer);

S. Senthil Kumar wrote:

Because I can't see how *anything* can be passed by reference at all - ultimately, an address would have to pushed on to the stack and popped at the other end.

Guffa wrote:

The specification only says that the value of the variable will be updated when the control passes back to the calling method, so the value of the variable could actually be passed back and forth instead of a passing a pointer to the variable.

struct S 
{
  void SayHello() {} 
}

void Caller()
{
   S s;
   Callee(s);
}

void Callee(S s)
{
   s.SayHello();
}

struct S 
{
  void SayHello() {} 
}

void Caller()
{
   S s;
   Callee(&s);
}

void Callee(S* s)
{
   s->SayHello();
}

DaveyM69 wrote:

ref has to be preassigned but does not have to be used in the method

DaveyM69 wrote:

out must be assigned before returning but can be unassigned before calling the method

DaveyM69 wrote:

Just discovered that if using out, the parameter is treated as if it is unassigned (even if it's already been assigned) as soon as you call the method.

arkiboys wrote:

Not sure where to check.