When to use reinterpret_cast?

Question

I am little confused with the applicability of reinterpret_cast vs static_cast. From what I have read the general rules are to use static cast when

Answer 1

template <class outType, class inType>
outType safe_cast(inType pointer)
{
    void* temp = static_cast<void*>(pointer);
    return static_cast<outType>(temp);
}

I tried to conclude and wrote a simple safe cast using templates. Note that this solution doesn't guarantee to cast pointers on a functions.

Answer 2

The C++ standard guarantees the following:

static_casting a pointer to and from void* preserves the address. That is, in the following, a, b and c all point to the same address:

int* a = new int();
void* b = static_cast<void*>(a);
int* c = static_cast<int*>(b);

reinterpret_cast only guarantees that if you cast a pointer to a different type, and then reinterpret_cast it back to the original type, you get the original value. So in the following:

int* a = new int();
void* b = reinterpret_cast<void*>(a);
int* c = reinterpret_cast<int*>(b);

a and c contain the same value, but the value of b is unspecified. (in practice it will typically contain the same address as a and c, but that's not specified in the standard, and it may not be true on machines with more complex memory systems.)

For casting to and from void*, static_cast should be preferred.

Answer 3

The meaning of reinterpret_cast is not defined by the C++ standard. Hence, in theory a reinterpret_cast could crash your program. In practice compilers try to do what you expect, which is to interpret the bits of what you are passing in as if they were the type you are casting to. If you know what the compilers you are going to use do with reinterpret_cast you can use it, but to say that it is portable would be lying.

For the case you describe, and pretty much any case where you might consider reinterpret_cast, you can use static_cast or some other alternative instead. Among other things the standard has this to say about what you can expect of static_cast (§5.2.9):

An rvalue of type “pointer to cv void” can be explicitly converted to a pointer to object type. A value of type pointer to object converted to “pointer to cv void” and back to the original pointer type will have its original value.

So for your use case, it seems fairly clear that the standardization committee intended for you to use static_cast.

Answer 4

You could use reinterprete_cast to check inheritance at compile time.
Look here: Using reinterpret_cast to check inheritance at compile time

Answer 5

Quick answer: use static_cast if it compiles, otherwise resort to reinterpret_cast.

Answer 6

The short answer: If you don't know what reinterpret_cast stands for, don't use it. If you will need it in the future, you will know.

Full answer:

Let's consider basic number types.

When you convert for example int(12) to unsigned float (12.0f) your processor needs to invoke some calculations as both numbers has different bit representation. This is what static_cast stands for.

On the other hand, when you call reinterpret_cast the CPU does not invoke any calculations. It just treats a set of bits in the memory like if it had another type. So when you convert int* to float* with this keyword, the new value (after pointer dereferecing) has nothing to do with the old value in mathematical meaning.

Example: It is true that reinterpret_cast is not portable because of one reason - byte order (endianness). But this is often surprisingly the best reason to use it. Let's imagine the example: you have to read binary 32bit number from file, and you know it is big endian. Your code has to be generic and works properly on big endian (e.g. some ARM) and little endian (e.g. x86) systems. So you have to check the byte order. It is well-known on compile time so you can write constexpr function: You can write a function to achieve this:

/*constexpr*/ bool is_little_endian() {
  std::uint16_t x=0x0001;
  auto p = reinterpret_cast<std::uint8_t*>(&x);
  return *p != 0;
}

Explanation: the binary representation of x in memory could be 0000'0000'0000'0001 (big) or 0000'0001'0000'0000 (little endian). After reinterpret-casting the byte under p pointer could be respectively 0000'0000 or 0000'0001. If you use static-casting, it will always be 0000'0001, no matter what endianness is being used.

EDIT:

_{In the first version I made example function is_little_endian to be constexpr. It compiles fine on the newest gcc (8.3.0) but the standard says it is illegal. The clang compiler refuses to compile it (which is correct).}