如何将char数组转换为字节数组?
char CardNumber[8] = "B763AB23"; // Length is 8, basically it's in Hex
// B7 63 AB 23
我需要将其转换为字节数组到byte CardNumberByte[4];
,因此基本上应该是:
CardNumberByte[0] = B7;
CardNumberByte[1] = 63;
CardNumberByte[2] = AB;
CardNumberByte[3] = 23;
我找不到任何解决方案。
union
{
uint32_t number;
uint8_t CardNumberByte[4];
} CardNum;
char cn[] = "B763AB23";
和"转换":
CardNum.Number = strtol(cn, NULL, 16);
,您的字节可通过
避免使用CardNum.CardNumberByte[xx]
,但我认为您应该从居民C 教程开始并学习一些基础知识。
八个十六进制字符是32位,因此首先将数字放入长(arduino上的32位(值:
long number = (long) strtol(&CardNumber[0], NULL, 16);
然后将值换到字节中:
CardNumberByte[0] = number >> 24;
CardNumberByte[1] = number >> 16 & 0xFF;
CardNumberByte[2] = number >> 8 & 0xFF;
CardNumberByte[3] = number & 0xFF;