这样的数据结构该怎样提取?
有这样一组数据:
44 54 00 07 9C A8 11 02 24 03 29 00
它的数据内容为:
44 54 00:DT, 07:07年, 9C A8:特征字;
11 02 24 03 29 00:110天,22时,40分,32.900秒
其中时间段的天、时分秒分别占用了下一个字节的8个位，我想用structure
定义一个结构变量：
type structure{
char id[3]; //DT
int yy; //07
char name[2] //9CA8
int jd; //110
int hh; //22
int mm; //40
float ss; //32.900
}
实际上这样定义只能提取出前2个字符（DT），提取其它的数据则完全错误，
该如何定义这个结构变量？请各位高人帮忙！先谢谢了！
[解决办法]
方法很多，下面的方法只是一个例子：
$ cat 0722_001.c
/*
* http://community.csdn.net/Expert/topic/5668/5668420.xml?temp=.2454492
*/
#include <stdio.h>
#include <string.h>
#define PRINT(x) printf( "%s = %s\n ", #x, x)
typedef struct {
unsigned char id[2 + 1]; /* DT */
unsigned char yy[2 + 1]; /* 07 */
unsigned char name[4 + 1]; /* 9CA8 */
unsigned char jd[3 + 1]; /* 110 */
unsigned char hh[2 + 1]; /* 40 */
unsigned char mm[2 + 1]; /* 40 */
unsigned char ss[6 + 1]; /* 32.900 */
} MYMTOM;
static unsigned char buf[] =
{
0x44, 0x54, 0x00, 0x07,
0x9C, 0xA8, 0x11, 0x02,
0x24, 0x03, 0x29, 0x00
};
int main(void)
{
MYMTOM x;
memset(&x, 0, sizeof(x));
memcpy(x.id, buf, sizeof(x.id));
sprintf(x.yy, "%02hhX ", buf[3]);
sprintf(x.name, "%02hhX%02hhX ", buf[4], buf[5]);
sprintf(x.jd, "%02hhX%.1hhX ", buf[6], buf[7] > > 4);
sprintf(x.hh, "%.1hhX%.1hhX ", buf[7] & 0x0F, buf[8] > > 4);
sprintf(x.mm, "%.1hhX%.1hhX ", buf[8] & 0x0F, buf[9] > > 4);
sprintf(x.ss, "%.1hhX%.1hhX.%.1hhX%02hhX ",
buf[9] & 0x0F, buf[10] > > 4,
buf[10] & 0x0F, buf[11]);
PRINT(x.id);
PRINT(x.yy);
PRINT(x.name);
PRINT(x.jd);
PRINT(x.hh);
PRINT(x.mm);
PRINT(x.ss);
return 0;
}
$ make CFLAGS=-Wall run
cc -Wall 0722_001.c -o 0722_001
./0722_001
x.id = DT
x.yy = 07
x.name = 9CA8
x.jd = 110
x.hh = 22
x.mm = 40
x.ss = 32.900
$
