Working out a data type?

Ok, with overloading, C++ checks the type of variable that was passes, and makes a call to that function with thoes operators.

But outside the scope of overloading, can you figure out what type of variable something is? Say I have a function that takes a void pointer - assuming it was passed a char* array (You can do that right?);

Code:

void AFunction(void *AnyValue) {
 SHORT x;

 if(GetType?(AnyValue) == Constant_Meaning_Of_Char)
  for(x=0; x < (sizeof(AnyValue) / sizeof(GetType(AnyValue))); x++)
  cout << AnyValue[0]
 else
  cout << GetType(AnyValue);
}

The idea being you can find the type that was passed to the function (Even if it wouldent work in this code). Does c++ have the ability to do this? And, will this return correctly if I pass it my own object? Say;

Code:

class ti {
 private:
  int One;
  int Two;
 public;
  ti() {One=0; Two=0;)
  ti(int one, int two) {One=one; Two=two;}
};

Would the function in question return type ti?

Thanks!

Have you considered using C++ templates?

EDIT:
Just in case you really do want to find the types, have you considered RTTI?

Havent heard of either of thoes :/.

Havent heard of either of thoes :/.

You can search the Web, articles, tutorials and examples should abound.

Actually, what problem are you really trying solve?

I have a function that takes two void pointers and attempts to cast them based on their type.

I have a function that takes two void pointers and attempts to cast them based on their type.

Why do you need to do that? Also, I think the use of void* causes the type information to be lost.

I just want to have one typecast.

Code:

 int x;
 char y;
 string I;

 cast(&x, &I);
 cast(&x, &y);

Code:

 int x;
 char y;
 string I;

 I=IntegerToString(x);
 y=IntegerToChar(x);

You could overload it I supose, but it would still be nice to know if its possible to retrive a type.

You can use typeid(), but in this case it looks like the use of stringstreams as per your other thread is more appropriate.

Alright, google has a nice tutorial on that, thank you again!