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Macro to put data in Little Endian

Posted on 2007-11-30
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Last Modified: 2010-04-21
My enivroment is Unix, C.  I get the following compile error on attached 'C' code.  Please advise.

"put_little_u32.c", line 8: error 4068: Missing comma seperator between par
ameters in macro "endian_put_native_to_little_u32".
#include <stdio.h>
 
/* This macro reverses data order so LSB is first. */
#define endian_put_native_to_little_u32(long src, char * dest) \
        dest[0] = (char)(src); \
	dest[1] = (char)(src >> 8); \
	dest[2] = (char)(src >> 16); \
	dest[3] = (char)(src >> 24);  /* line 8 */
 
 
main()     
{
   char data[4];  
   
   long pgn = 0x33221100;
   
   endian_put_native_to_little_u32(pgn, &data[0]);
   
   printf("data[0] = %d.\n", data[0]);
   printf("data[1] = %d.\n", data[1]);
   printf("data[2] = %d.\n", data[2]);
   printf("data[3] = %d.\n", data[3]);
}

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Question by:naseeam
7 Comments
 
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Expert Comment

by:Jaime Olivares
ID: 20386568
You just need to use as:
   endian_put_native_to_little_u32(pgn, data);
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Expert Comment

by:Jaime Olivares
ID: 20386779
does it work?
0
 
LVL 84

Accepted Solution

by:
ozo earned 360 total points
ID: 20387262
#include <stdio.h>

/* This macro reverses data order so LSB is first. */
#define endian_put_native_to_little_u32(src, dest) \
        (dest)[0] = (char)(src); \
        (dest)[1] = (char)((src) >> 8); \
        (dest)[2] = (char)((src) >> 16); \
        (dest)[3] = (char)((src) >> 24);  /* line 8 */


main()
{
   char data[4];

   long pgn = 0x33221100;

   endian_put_native_to_little_u32(pgn, &data[0]);

   printf("data[0] = %d.\n", data[0]);
   printf("data[1] = %d.\n", data[1]);
   printf("data[2] = %d.\n", data[2]);
   printf("data[3] = %d.\n", data[3]);
}
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Expert Comment

by:fridom
ID: 20387307
Maybe you are looking for htonl or ntohl ?

Regards
Friedrich
0
 
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Assisted Solution

by:Infinity08
Infinity08 earned 140 total points
ID: 20390883
>> "put_little_u32.c", line 8: error 4068: Missing comma seperator between parameters in macro "endian_put_native_to_little_u32".

macro's do not take arguments like functions do ... Macro arguments are typeless, so you don't have to specify the types of the arguments. They just get treated as a string, and are literally replaced in the macro body.

See ozo's corrections.

Also, take a look at fridom's suggestion ! ;)
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Expert Comment

by:ozo
ID: 20391805
Also, if you pass &data[0] rather than data
&data[0][1] would be parsed as &(data[0][1])
and taking [0][1] of a char[4] makes no sense, so it has to be
(&data[0])[1]
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Author Closing Comment

by:naseeam
ID: 31412065
Expert Fixed my source code.  As a result my code compiled.
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