Practical Use of Zero-Length Bitfields

Question

I am not totally sure about C, but C++ allows unnamed bit-fields of 0 length. For example:

struct X {     int : 0; }; 

• Question one: What practical uses of this can you think of?
• Question two: What real-world practical uses (if any) are you aware of?

Edited the example after ice-crime's answer

Edit: OK, thanks to the current answers I now know the theoretical purpose. But the questions are about practical uses so they still hold :)

Answer 1

You use a zero-length bitfield as a hacky way to get your compiler to lay out a structure to match some external requirement, be it another compiler's or architecture's notion of the layout (cross-platform data structures, such as in a binary file format) or a bit-level standard's requirements (network packets or instruction opcodes).

A real-world example is when NeXT ported the xnu kernel from the Motorola 68000 (m68k) architecture to the i386 architecture. NeXT had a working m68k version of their kernel. When they ported it to i386, they found that the i386's alignment requirements differed from the m68k's in such a way that an m68k machine and an i386 machine did not agree on the layout of the NeXT vendor-specific BOOTP structure. In order to make the i386 structure layout agree with the m68k, they added an unnamed bitfield of length zero to force the NV1 structure/nv_U union to be 16-bit aligned.

Here are the relevant parts from the Mac OS X 10.6.5 xnu source code:

/* from xnu/bsd/netinet/bootp.h */ /*  * Bootstrap Protocol (BOOTP).  RFC 951.  */ /*  * HISTORY  *  * 14 May 1992 ? at NeXT  *  Added correct padding to struct nextvend.  This is  *  needed for the i386 due to alignment differences wrt  *  the m68k.  Also adjusted the size of the array fields  *  because the NeXT vendor area was overflowing the bootp  *  packet.  */ /* . . . */ struct nextvend {   u_char nv_magic[4]; /* Magic number for vendor specificity */   u_char nv_version;  /* NeXT protocol version */   /*    * Round the beginning    * of the union to a 16    * bit boundary due to    * struct/union alignment    * on the m68k.    */   unsigned short  :0;   union {     u_char NV0[58];     struct {       u_char NV1_opcode;  /* opcode - Version 1 */       u_char NV1_xid; /* transcation id */       u_char NV1_text[NVMAXTEXT]; /* text */       u_char NV1_null;  /* null terminator */     } NV1;   } nv_U; }; 

Answer 2

The standard (9.6/2) only allows 0 length bit-fields as a special case :

As a special case, an unnamed bit-field with a width of zero specifies alignment of the next bit-field at an allocation unit boundary. Only when declaring an unnamed bit-field may the constant-expression be a value equal to zero.

The only use is described in this quote, although I've never encountered it in practical code yet.

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For the record, I just tried the following code under VS 2010 :

struct X {     int i : 3, j : 5; };  struct Y {     int i : 3, : 0, j : 5; // nice syntax huh ? };  int main() {     std::cout << sizeof(X) << " - " << sizeof(Y) << std::endl; } 

The output on my machine is indeed : 4 - 8.