static void Main(string[] args)
{
MyMutableType mut = new MyMutableType(2);
double d = MyMutableType.BadFunc(mut);
Console.WriteLine(mut.x);
//The console will print out '4' instead of '2', which was the value of x before the call to BadFunc.
// Our object's data has changed, and we wouldn't even know it, if we didn't write BadFunc ourselves.
// The reason is, that when we pass m to BadFunc, we do not create a copy of the object m, but only of the reference to the object m.
// Now m and the local copy t both reference the same object, and inside BadFunc, we can manipulate this object,
// and the changes are visible globally, as we can see when printing the value of x after the call to BadFunc
MyImmutableType m = new MyImmutableType(2);
MyImmutableType.TriesToBeBadFunc(m);
Console.WriteLine(m.x);
// The console will print out '2'.
// There are no methods anymore in MyImmutableType which could change its state or data,
// so m cannot be manipulated through t anymore, and the assignment t = t.Square() inside the method simply assigns the new instance created by t.Square() to t,
// but because t is a copy of the reference m, this assignment only changes the reference t, and m stays unchanged.
}
public static double BadFunc(MyMutableType t)
{
t.Square();
return(t.x);
}
