c - 如何在不在 C 中使用 union 的情况下将十六进制字节数组连接到 long int？

Question

我有一个数组。数组字节是十六进制的。现在我试图将数组索引 1 到 4 连接到 unsigned long int，将 5 到 8 连接到另一个 unsigned long int，将索引 9 和 10 连接到 unsigned int。我试过这样。但它不起作用。我不想使用 union 来连接。我在 STM32F4-Discovery 中使用此代码。

这是一个接收中断处理函数:

  void usart1_IRqHandler()
  {
  unsigned char received_string[20];
  unsigned long int Addr1=0,Addr2=0;
  unsigned int Addr=0;
  
  /*Code to Receive Entire packet.......
   .....................................
   .....................................
  At this point whole packet is received, now I want to send back the response to the target node using address from received data, So first step is to concatenate address bytes*/ 
  
  /*From the logic analyzer I can observe the received_string bytes as shown below 
  received_string[1]=0x00; 
  received_string[2]=0x13;
  received_string[3]=0xA2;
  received_string[4]=0x00;

  received_string[5]=0x40;
  received_string[6]=0xB4;
  received_string[7]=0x14;
  received_string[8]=0x35;

  received_string[9]=0x8E;
  received_string[10]=0xC7;*/

  Addr1 =  (received_string[1]<<24)| (received_string[2]<<16)| (received_string[3]  <<8)| received_string[4]);
  Addr2 =  (received_string[5]<<24)| (received_string[6]<<16)|(received_string[7]  <<8)| received_string[8]);           
  Addr =  (received_string[9]<<8)| (received_string[10]);   

  Send_packet(Addr1,Addr2,Addr);
  }

函数发送数据包将以这些地址为目标远程节点。

     Send_packet(unsigned long int Addr1,unsigned long int Addr1,unsigned int Addr)
    {
      unsigned char tx_buf[20];
      tx_buf[5]=(Addr1>>24) & 0xff;
      tx_buf[6]=(Addr1>>16) & 0xff;
      tx_buf[7]=(Addr1>>8)  & 0xff;
      tx_buf[8]=(Addr1>>0)  & 0xff;

      tx_buf[9]=(Addr2>>24)  & 0xff;
      tx_buf[10]=(Addr2>>16) & 0xff;
      tx_buf[11]=(Addr3>>8)  & 0xff;
      tx_buf[12]=(Addr4>>0)  & 0xff;

      tx_buf[9]=(Addr2>>8)  & 0xff;
      tx_buf[10]=(Addr2>>0) & 0xff;
     
      usart_send(tx_buf);
      
   }

这是逻辑分析仪发送数据包的输出。

  Addr1=0xA2401437
  Addr2=0x025500A2
  Addr=0x4014

由于地址不匹配，我无法定位到远程节点。

Answer 1

问题缺少一个 minimal reproducable example并且缺乏有关如何获取信息的详细信息。但是，如何转换数字的基本思路是正确的(问题中的代码无法编译，因此无法判断 OP 实际测试了哪些代码)。

下面是一个演示其工作原理的示例:

#include <stdio.h>

int main()
{
    unsigned char received_string[20];
    
    received_string[1] = 0x00;
    received_string[2] = 0x13;
    received_string[3] = 0xA2;
    received_string[4] = 0x00;
    
    received_string[5] = 0x40;
    received_string[6] = 0xB4;
    received_string[7] = 0x14;
    received_string[8] = 0x35;

    received_string[9] = 0x8E;
    received_string[10] = 0xC7;
    
    unsigned long int Addr1 = (received_string[1] << 24) | (received_string[2] << 16) | 
                              (received_string[3] << 8) | received_string[4];
    unsigned long int Addr2 = (received_string[5] << 24) | (received_string[6] << 16) | 
                              (received_string[7] << 8) | received_string[8];
    unsigned int Addr = (received_string[9] << 8) | (received_string[10]);
    
    printf("Addr1 = %08lX\n", Addr1);
    printf("Addr2 = %08lX\n", Addr2);
    printf("Addr = %04X\n", Addr);

    return 0;
}

这将打印:

Addr1 = 0013A200
Addr2 = 40B41435
Addr = 8EC7

注意:正如用户 M.M. 所正确陈述的那样。在comment在 answer of Yasir Majeed 上，掩蔽是完全没有必要的。它要么被编译器优化掉，要么插入无用的指令。

但是请注意，为了使此代码更具可移植性，应将字节转换为 unsigned long在转移之前。否则，这将在 sizeof(int) < 4 所在的平台上失败。 .