怎么将一字符串动态赋值成16进制值，内含详细代码（总共几行）

如何将一字符串动态赋值成16进制值，内含详细代码（总共几行）
MDK环境下，有如下实现

uint8_t Settimebuf[7];

Settimebuf[6] = 0x12; //年
Settimebuf[5] = 0x07; //周
Settimebuf[4] = 0x06; //月
Settimebuf[3] = 0x17; //日
Settimebuf[2] = 0x10; //时
Settimebuf[1] = 0x23; //分
Settimebuf[0] = 0x20;//秒

Set1381Time(Settimebuf);

现有一个char变量动态记录了年月日时分秒的信息"120617102320"，请教如何将字符串中相应时间信息赋值成如上，以实现可动态设置时间信息

谢谢，一个非C程序员

[解决办法]
不知道你为什么要用16进制，如果一定要用的话，那你就改一处代码即可，把×10 改成×16

C/C++ code

#include <stdio.h>#define GET_NUM(ch) ((ch) - '0')unsigned char* set_buf(const char* str, unsigned char* buf, const size_t LEN) {    size_t i;    unsigned char num = 0;    for (i = 0; i < LEN; ++i) {        num = GET_NUM(str[2*i]) * 10 + GET_NUM(str[2*i+1]);        buf[i] = num;    }    return buf;}/* 测试代码 */int main() {    const char *str = "120617102320";    size_t i;    unsigned char buf[7];    const size_t LEN = sizeof(buf);    set_buf(str, buf, 7);    for (i = 0; i < LEN; ++i) {        printf("%d\n", buf[i]);    }    return 0;}
[解决办法]
C/C++ code
[User:root Time:21:23:37 Path:/home/liangdong/c]$ makegcc -g -I./include   -c -o src/main.o src/main.cgcc -o output src/main.o -lpthread -lm -lzMakefile done.[User:root Time:21:23:37 Path:/home/liangdong/c]$ ./output 0x12 0x7 0x6 0x17 0x10 0x23 0x20 [User:root Time:21:23:38 Path:/home/liangdong/c]$ cat src/main.c #include <stdio.h>#include <stdlib.h>#include <string.h>#include <stdint.h>int strpuint8(const char *time, uint8_t *buf) {        if (!time) {                return -1;        }        int i;        for (i = 0; i != 7; ++ i) {                if (sscanf(time + i * 2, "%2x", buf + i) != 1) {                        return -1;                }        }        return 0;}int main(int argc, char* const argv[]) {        uint8_t set_time_buf[7];        const char *time = "12070617102320";        int ret = strpuint8(time, set_time_buf);        if (ret == 0) {                int i;                for (i = 0; i != 7; ++ i) {                        printf("0x%x ", set_time_buf[i]);                }        }        printf("\n");        return 0;}