Returning different data type depending on the data (C++)

Question

Is there anyway to do something like this?

(correct pointer datatype) returnPointer(void* ptr, int depth)
{

    if(depth == 8)
        return (uint8*)ptr;
    else if (depth == 16)
        return (uint16*)ptr;
    else
        return (uint32*)ptr;
}

Thanks

Answer 1

If you can use a template argument instead of a normal parameter, you can create a templated function that returns the correct type for each depth value. First there needs to be some definition of the correct type according to depth. You can define a template with specializations for the different bit sizes:

// template declaration
template<int depth>
struct uint_tmpl;

// specializations for certain types
template<> struct uint_tmpl<8>  { typedef uint8_t type; };
template<> struct uint_tmpl<16> { typedef uint16_t type; };
template<> struct uint_tmpl<32> { typedef uint32_t type; };

The this definition can be used to declare a templated function that returns the correct type for every bit value:

// generic declaration
template<int depth>
typename uint_tmpl<depth>::type* returnPointer(void* ptr);

// specializations for different depths
template<> uint8_t*  returnPointer<8>(void* ptr)  { return (uint8_t*)ptr;  }
template<> uint16_t* returnPointer<16>(void* ptr) { return (uint16_t*)ptr; }
template<> uint32_t* returnPointer<32>(void* ptr) { return (uint32_t*)ptr; }

Answer 2

No. The return type of a C++ function can only vary based on explicit template parameters or the types of its arguments. It cannot vary based on the value of its arguments.

However, you can use various techniques to create a type that is the union of several other types. Unfortunately this won't necessarily help you here, as one such technique is void * itself, and getting back to the original type will be a pain.

However, by turning the problem inside out you may get what you want. I imagine you'd want to use the code you posted as something like, for example:

void bitmap_operation(void *data, int depth, int width, int height) {
  some_magical_type p_pixels = returnPointer(data, depth);
  for (int x = 0; x < width; x++)
    for (int y = 0; y < width; y++)
      p_pixels[y*width+x] = some_operation(p_pixels[y*width+x]);
}

Because C++ needs to know the type of p_pixels at compile time, this won't work as-is. But what we can do is make bitmap_operation itself be a template, then wrap it with a switch based on the depth:

template<typename PixelType>
void bitmap_operation_impl(void *data, int width, int height) {
  PixelType *p_pixels = (PixelType *)data;
  for (int x = 0; x < width; x++)
    for (int y = 0; y < width; y++)
      p_pixels[y*width+x] = some_operation(p_pixels[y*width+x]);
}

void bitmap_operation(void *data, int depth, int width, int height) {
  if (depth == 8)
    bitmap_operation_impl<uint8_t>(data, width, height);
  else if (depth == 16)
    bitmap_operation_impl<uint16_t>(data, width, height);
  else if (depth == 32)
    bitmap_operation_impl<uint32_t>(data, width, height);
  else assert(!"Impossible depth!");
}

Now the compiler will automatically generate three implementations for bitmap_operation_impl for you.

Answer 3

You can allocate some memory on the heap, and return a void* that you cast to the type that was allocated. Its a dangerous and unsafe way of working and is an old C trick.

You could return a union that contains all valid datatypes (and a selection indicator).

You could use templates, which is the recommended C++ way for this kind of thing.

You could provide a set of overloaded functions that take a parameter (of each type) as a reference - the compiler will decide which function to call based on the datatype. I often prefer this way as I think its the simplest.