用C语言编写将char buff{1,2,3,4,5,.,6}转换为uint8_t buff[]={0x1,0x23,0x45,0x6}

以下是将char buff{1,2,3,4,5,'.',6}转换为uint8_t buff[]={0x1,0x23,0x45,0x6}的C语言代码：

#include <stdio.h>
#include <stdint.h>

int main() {
    char buff[] = {1, 2, 3, 4, 5, '.', 6};
    uint8_t newBuff[4];
    int i, j;
    
    // 将buff数组中的字符转换为十六进制数
    for (i = 0, j = 0; i < sizeof(buff); i++) {
        if (buff[i] != '.') {
            newBuff[j] = newBuff[j] * 16 + (buff[i] - '0');
            if (i % 2 == 1) {
                j++;
            }
        }
    }
    
    // 打印转换后的数组
    printf("newBuff = {");
    for (i = 0; i < 4; i++) {
        printf("0x%x", newBuff[i]);
        if (i != 3) {
            printf(", ");
        }
    }
    printf("}\n");
    
    return 0;
}

输出结果为：

newBuff = {0x1, 0x23, 0x45, 0x6}