About static_cast in c++

Question

I am reading http://www.gotw.ca/gotw/036.htm (it is about direct initialization and copy initialization) and it has a mention like:

static_cast<S>(t); // performing a static_cast

A static_cast uses direct initialization.

Can someone explain why and how does static_cast need to use initialization? I thought that casting is just reinterpreting the bits in a given memory location...

Casting is not just reinterpreting bits though. Casting from `int` to `float` changes the bits in the memory. — rozina, Mar 21 '14 at 11:17
possible duplicate of [When should static\_cast, dynamic\_cast, const\_cast and reinterpret\_cast be used?](http://stackoverflow.com/questions/332030/when-should-static-cast-dynamic-cast-const-cast-and-reinterpret-cast-be-used) — Shoe, Mar 21 '14 at 11:28

score 5 · Answer 1 · answered Mar 21 '14 at 11:18

A reinterpret_cast reinterprets the bits. A static_cast is a semantics-aware conversion. It will try to find a sensible operation that converts a value from one type to another, and fail to compile if it can't.

Sensible operations are conversion (possibly with truncation) between various integer and floating point types, automatic conversions to bool, converting bool to numbers, etc. And also, using user-defined conversion operators and single-argument constructors to convert from and to user-defined types. And in the case where it uses constructors, it uses direct-initialization, and thus can use explicit constructors, unlike implicit conversion, which cannot.

score 0 · Answer 2 · answered Mar 21 '14 at 11:23

static_cast is performing look up in the class hierarchy in order to cast to the correct class, it doesn't reinterpret the bits (that is reinterpret_cast).

The same thing happens with dynamic_cast, the difference being that static_cast is done at compile time while dynamic_cast is done at runtime.

Take the following example : A : B , A : C , D : B,C

The following code will print that the pointers point to a different memory block

{
  D *ptrToD = new D();
  C *ptr1 = static_cast<C>(ptrToD);
  C *ptr2 = reinterpret_cast<C>(ptrToD);
  if (ptr1==ptr2)
  {
     printf("Pointers point to the same memory block");
  } else
  {
     printf("Pointers point to different memory blocks");
  }
}

If you look at the memory , it will look something like ABACD

If you want to cast it to C , C starts from where D starts +sizeof(B) (AC) ,whereas A and B starts from the same memory block as D.

score 0 · Answer 3 · answered Mar 21 '14 at 11:29

Maybe this is not an fully explanatory example, but watch the following

void f(double **arg) {} // just to make a point

int main() {
    double d(1.);
    double *pd = &d;
    double const *cpd = &d;

    f(&pd); // OK 
    f(&(const_cast<double*>(cpd))); // Error
}

The emmited error is

error C2102: & requires l-value

The static_cast would behave the same way, in the sense that they return an r-value which you must bind to a variable. It is that variable that gets initialized

score 0 · Answer 4 · answered Mar 21 '14 at 11:33

I thought that casting is just reinterpreting the bits in a given memory location...

No, casting performs a type conversion. In this case, it converts t to type S; such a conversion is possible since T inherits from S, so a new S can be made by copying that part of t. Copying a base sub-object like that is sometimes known as slicing.

Can someone explain why and how does static_cast need to use initialization?

A type conversion requires the creation of a (temporary) object of the new type; that object needs to be initialised and the rules for static_cast specify that it use direct initialisation. Since S is a class type, this means it's initialised by calling a suitable constructor - in this case, the implicitly-declared copy constructor, which initialises the object by copying the S sub-object of t.

About static_cast in c++

4 Answers4