Is it possible to pass an integer as reference at class initialization and safe the reference?
class Foo {
private int _refVal;
public Foo(ref int val) {
_refval = val; // saves the value, not the reference
}
}
I could use pointers, but then I need an unsafe context.
This is not possible.
Instead, you can use a class with a writable property, like this:
class Reference<T> {
public T Value { get; set; }
public Reference(T value) { Value = value; }
}
Out of interest, why do you need to do this? One integer equal to 5 is equal to another integer equal to 5: if there is some differentiation you want to make between them, the integer value type shouldn't be used - you'd want a class instead.
This is not a direct answer to your question, but as they say improving an algorithm is better than implementing or improving a flawed one; perhaps if you could give us some more context we can help with your more general problem / task as a whole?
Hope that helps!
Wrap it in a custom class I guess.
