Set a FourCC value in C++

Question

I want to set a FourCC value in C++, i.e. an unsigned 4 byte integer.

I suppose the obvious way is a #define, e.g.

#define FOURCC(a,b,c,d) ( (uint32) (((d)<<24) | ((c)<<16) | ((b)<<8) | (a)) )

and then:

uint32 id( FOURCC('b','l','a','h') );

What is the most elegant way you can think to do this?

Answer 1

I see nothing wrong with your algorithm. But for something like this I would just write a function instead of a macro. Macros have a lot of hidden features / problems that can bite you over time.

uint FourCC(char a, char b, char c, char d) { 
  return ( (uint32) (((d)<<24) | ((c)<<16) | ((b)<<8) | (a)) );
}

Answer 2

or do the same with an inline function

inline uint32_t FOURCC(uint8_t a, uint8_t b, uint8_t c, uint8_t d)
{
     return ( (uint32) (((d)<<24) | (uint32_t(c)<<16) | (uint32_t(b)<<8) | uint32_t(a)) )
}

and avoid the headaches of a macro, but otherwise your approach looks fine to me.

Answer 3

Rather than a #define, I'd probably put pretty much the same code and rely on the compiler to inline it.

Answer 4

You can make it a compile-time constant using:

template <int a, int b, int c, int d>
struct FourCC
{
    static const unsigned int value = (((((d << 8) | c) << 8) | b) << 8) | a;
};

unsigned int id(FourCC<'a', 'b', 'c', 'd'>::value);

With a little extra effort, you can make it check at compile time that each number passed in is between 0 and 255.

Answer 5

If I am not mistaken, you can just use multi-character character constants for that right?

unsigned int fourCC = 'blah';

This is perfectly valid by the ANSI/ISO specification though some compilers will complain a little. This is how resource types used to be handled in the older Macintosh APIs.

Answer 6

Assuming Windows (as FOURCC is a Windows concept), the Win API already provides mmioStringToFOURCC and mmioFOURCC.

Answer 7

If a compile-time constant isn't required, perhaps the neatest is

unsigned int FourCCStr(const char (&tag)[5])
{
    return (((((tag[3] << 8 ) | tag[2]) << 8) | tag[1]) << 8) | tag[0];
}

#define FOURCC(tag) FourCCStr(#tag)

unsigned int id(FOURCC(blah));

This only accepts tags of four characters, as required.

Answer 8

uint32_t FourCC = *((uint32_t*)"blah");

Why not this?

EDIT: int -> uint32_t.

And no it does not cast a char** to uint32_t. It casts a (char*) to (uint32_t*) then dereferences the (uint32_t*). There is no endian-ness involved, since its assigning an uint32_tto an uint32_t. The only defects are the alignment and the I hadn't explicitly indicated a 32bit type.

Answer 9

uint32 fcc(char * a)
{   
    if( strlen(a) != 4)
     return 0;  //Unknown or unspecified format

    return 
    (
      (uint32) 
      ( 
       ((*(a+3))<<24) |
       ((*(a+2))<<16) |
       ((*(a+1))<<8) | 
       (*a)
      )  
    );
}