• jc (4/4) Nov 29 2007 hi,
• Bill Baxter (7/13) Nov 29 2007 Take the address, cast to byte, deref byte by byte:
• Regan Heath (5/20) Nov 30 2007 Which means that you can go:
• Jason House (4/21) Nov 30 2007 Does gdc run on any little endian machines? I'm not sure which way arra...
• Bill Baxter (5/27) Nov 30 2007 Yep. Order most definitely depends on endianness. Intel is little
• Stewart Gordon (12/15) Dec 06 2007 "Jason House" wrote in message

jc <jc dummy.com> writes:

hi,

is there a way to stuff the 8 bytes of a long into a byte array other than to
use a union to achive that?

cheers
jc

Bill Baxter <dnewsgroup billbaxter.com> writes:

jc wrote:
 hi,
 
 is there a way to stuff the 8 bytes of a long into a byte array other than to
use a union to achive that?
 
 cheers
 jc

Take the address, cast to byte, deref byte by byte:

     (cast(ubyte*)&the_long)[0];
     (cast(ubyte*)&the_long)[1];
     (cast(ubyte*)&the_long)[2];
     ...

--bb

Regan Heath <regan netmail.co.nz> writes:

Bill Baxter wrote:
 jc wrote:
 hi,

 is there a way to stuff the 8 bytes of a long into a byte array other 
 than to use a union to achive that?

 cheers
 jc

 
 Take the address, cast to byte, deref byte by byte:
 
     (cast(ubyte*)&the_long)[0];
     (cast(ubyte*)&the_long)[1];
     (cast(ubyte*)&the_long)[2];
     ...

Which means that you can go:

byte[] array = (cast(ubyte*)&the_long)[0..8];

if you want an actual array.

Regan

Jason House <jason.james.house gmail.com> writes:

Bill Baxter wrote:

 jc wrote:
 hi,
 
 is there a way to stuff the 8 bytes of a long into a byte array other
 than to use a union to achive that?
 
 cheers
 jc

 
 Take the address, cast to byte, deref byte by byte:
 
      (cast(ubyte*)&the_long)[0];
      (cast(ubyte*)&the_long)[1];
      (cast(ubyte*)&the_long)[2];
      ...
 
 --bb

Does gdc run on any little endian machines?  I'm not sure which way arrays
grow on such machines, but I think this could would give the reverse byte
order of a big endian machine.

Bill Baxter <dnewsgroup billbaxter.com> writes:

Jason House wrote:
 Bill Baxter wrote:
 
 jc wrote:
 hi,

 is there a way to stuff the 8 bytes of a long into a byte array other
 than to use a union to achive that?

 cheers
 jc

 Take the address, cast to byte, deref byte by byte:

      (cast(ubyte*)&the_long)[0];
      (cast(ubyte*)&the_long)[1];
      (cast(ubyte*)&the_long)[2];
      ...

 --bb

 
 Does gdc run on any little endian machines?  I'm not sure which way arrays
 grow on such machines, but I think this could would give the reverse byte
 order of a big endian machine.

Yep.  Order most definitely depends on endianness.  Intel is little 
endian.  If you're planning on sending those bytes over a network you'd 
better swap em if you've got version(LittleEndian).

--bb

"Stewart Gordon" <smjg_1998 yahoo.com> writes:

"Jason House" <jason.james.house gmail.com> wrote in message 
news:fipgc3$5ni$1 digitalmars.com...
<snip>
 Does gdc run on any little endian machines?  I'm not sure which way arrays
 grow on such machines, but I think this could would give the reverse byte
 order of a big endian machine.

Array indexes always follow the sign convention of memory addresses.  The 
basic difference between big-endian and little-endian machines is which way 
the bytes of a number go relative to this sign convention.  So yes.

As such, accessing an integer as an array of bytes is a common technique for 
converting between little-endian and big-endian byte orders.

Stewart.

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