将 MFRC522 UID 十六进制字节转换为可打印的十进制 | 十六进制字节转换为可打印的十进制

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问题描述

我在我的 Arduino UNO 上使用 MFRC522 库来读取 Mifare RFID 标签信息.>

I'm using the MFRC522 library on my Arduino UNO to read Mifare RFID tag Info.

// Print HEX UID
Serial.print("Card UID:");
for (byte i = 0; i < mfrc522.uid.size; i++) {
    Serial.print(mfrc522.uid.uidByte[i] < 0x10 ? " 0" : " ");
    Serial.print(mfrc522.uid.uidByte[i], HEX);
}
Serial.println();

我有一个包含十六进制 UID 的字节数组 (4):

I've got a byte array(4) which contains the HEX UID:

54 C7 FD 5A

但我未能将其转换为十进制:

But I've failed to convert it to Decimal:

HEX(5AFDC754) => DEC(1526581076)

我尝试将字节数组反向转换为字符，但编译器不让我打印 12 月.

I've tried to convert the byte array to char reversely, but the compiler didn't let me print Dec.

char str[8];
int k = 0;

for (int i = 3; i >= 0 ; i -= 1) {
    char hex[4];
    snprintf(s, 4, "%x", mfrc522.uid.uidByte[i]);

    for( int t = 0; t < 4; t++ ) {
        if( (int)hex[t] != 0 )
            str[t+k] = hex[t];
    }

    k+=2;
}

Serial.println( str, DEC);

感谢任何建议

推荐答案

您需要将 4 个十六进制字节组合成一个无符号整数.

You will need to combine the 4 hex bytes into a single unsigned integer.

这取决于 Endianess(搜索它).

This depends on Endianess (search for it).

对于大端:

  unsigned int hex_num;
  hex_num =  uidByte[0] << 24;
  hex_num += uidByte[1] << 16;
  hex_num += uidByte[2] <<  8;
  hex_num += uidByte[3];

对于 Little Endian，颠倒 uidByte 位置的顺序.

For Little Endian, reverse the order of uidByte positions.

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