Convert the raw data in little endian format on x64 system
The raw data gets from the device in endian neutral format. I want to build a common subroutine to convert it in little endain before to copy data out in a structure (don't care how the structure is defined).
For example, a 16-byte of data from the device
Byte_0 _1 _2 _3 _4 _5 _6 _7 _8 _9 _A _B _C _D _E _F
-
Data00 01 00 01 10 00 00 17 00 10 00 01 01 00 00 02
is converted in this order of little endian.
Byte_3 _2 _1 _0 _7 _6 _5 _4 _B _A _9 _8 _F _E _D _C
-
Data01 00 01 00 17 00 00 10 01 00 10 00 02 00 00 01
If the structure is defined as the following:
struct {
short A;/* Byte 0-1 */
short B;/* Byte 2-3 */
short C;/* Byte 4-5 */
short D;/* Byte 6-7 */
short E;/* Byte 8-9 */
short F;/* Byte A-B */
short G; /* Byte C-D */
short H; /* Byte E-F */
}
The printout is
Short A ... 0x1/* Byte 2-3 */
Short B ... 0x1/* Byte 0-1 */
Short C ... 0x 17/* Byte 6-7 */
Short D ... 0x1000 /* Byte 4-5 */
Short E ... 0x1/* Byte A-B */
Short F ... 0x 10/* Byte 8-9 */
Short G ... 0x2/* Byte E-F */
Short H ... 0x 100 /* Byte C-D */
The source code is post here.
void byte_convert (char *data, uint size)
{
int i=0, j;
char *tmp;
(void) memset (tmp, 0, size);
for (j=0; j<size; j=j+4) {
switch(size%4) {
case 0:
*(tmp+i)= *(data+i+3);
*(tmp+i+1) = *(data+i+2);
*(tmp+i+2) = *(data+i+1);
*(tmp+i+3) = *(data+i);
i = i + 4;
break;
case 1:
*(tmp+i)= *(data+i);
i = i + 1;
break;
case 2:
*(tmp+i)= *(data(i+1);
*(tmp+i+1) = *(data(i);
i = i + 2;
break;
case 3:
*(tmp+i)= *(data+i+2);
*(tmp+i+1) = *(data+i+1);
*(tmp+i+2) = *(data+i);
i = i + 3;
break;
}
}
bcopy(tmp, data, size);
}
return;
}
How is the data located in the memory on Solaris x64 system? What is the wrong in my code?>
