C语言编程中的8位、16位、32位整数的分解与合并

一时失言乱红尘 2022-01-11 17:45 934阅读 0赞

在单片机的编程中对于8位、16位、32位整数的分解与合并用的比较多，今天做了简要学习，后面还需要加以总结。

练习在VC++6.0编程环境中进行，源程序：  
\#include <stdio.h>  
\#include "string.h"

int main(int argc, char *argv\[\])  
\{  
unsigned int Data\_Uint32=0x12345678;  
unsigned short int Data\_Uint16\_1,Data\_Uint16\_2;  
unsigned char Data\_Uint8\_1,Data\_Uint8\_2,Data\_Uint8\_3,Data\_Uint8\_4;  
struct StructByte8\{  
unsigned char Byte01:1;  
unsigned char Byte02:1;  
unsigned char Byte03:1;  
unsigned char Byte04:1;  
unsigned char Byte05:1;  
unsigned char Byte06:1;  
unsigned char Byte07:1;  
unsigned char Byte08:1;  
\} Test1;  
unsigned char C1\[\]="A";  
unsigned short int* p16=(unsigned short int \*)(&Data\_Uint32);//定义16位的指针将32位地址强制转为16位，高位丢弃取低位  
unsigned short int Data\_Uint16\_2p,Data\_Uint16\_1p;

printf("32位整数：0x%x\n",Data_Uint32);
      printf("-------------------通过指针运算-------------------\n");
      Data_Uint16_2p=*(unsigned short int *)p16;
      Data_Uint16_1p=*((unsigned short int *)p16+1);
      printf("转换后的16位整数：0x%x,0x%x\n",Data_Uint16_1p,Data_Uint16_2p);
    
      Data_Uint8_1= *(unsigned char *)p16;
      Data_Uint8_2= *((unsigned char *)p16+1);
      Data_Uint8_3= *((unsigned char *)p16+2);
      Data_Uint8_4= *((unsigned char *)p16+3);
      printf("转换后的8位整数：0x%x,0x%x,0x%x,0x%x\n",Data_Uint8_1,Data_Uint8_2,Data_Uint8_3,Data_Uint8_4);
    
      Data_Uint8_1=0;Data_Uint8_2=0;Data_Uint8_3=0;Data_Uint8_4=0;
      //直接根据指针取值
      //将32位的整数分解成两个16位的整数,再取低位的16位
      //强制转换，丢弃高位的16位
      //32位转16位
      Data_Uint16_1=(unsigned short int)(Data_Uint32>>16);
      Data_Uint16_2=(unsigned short int)Data_Uint32;
    
      //32位转8位
      Data_Uint8_1= (unsigned char)(Data_Uint32>>24);
      Data_Uint8_2= (unsigned char)(Data_Uint32>>16);
      Data_Uint8_3= (unsigned char)(Data_Uint32>>8);
      Data_Uint8_4= (unsigned char)Data_Uint32;
    
      printf("-------------------通过位运算-------------------\n");
      printf("转换后的16位整数：0x%x,0x%x\n",Data_Uint16_1,Data_Uint16_2);
      printf("转换后的8位整数：0x%x,0x%x,0x%x,0x%x\n",Data_Uint8_1,Data_Uint8_2,Data_Uint8_3,Data_Uint8_4);
    
      printf("-------------------通过结构运算-------------------\n");     
      printf("字符A，ASCII为01000001，结构转换后的内容\n");  
      memcpy(&Test1, C1, sizeof(Test1));  
      printf("Test1.Byte08=%d \n",Test1.Byte08);  
      printf("Test1.Byte07=%d \n",Test1.Byte07);  
      printf("Test1.Byte06=%d \n",Test1.Byte06);  
      printf("Test1.Byte05=%d \n",Test1.Byte05);  
      printf("Test1.Byte04=%d \n",Test1.Byte04);  
      printf("Test1.Byte03=%d \n",Test1.Byte03);  
      printf("Test1.Byte02=%d \n",Test1.Byte02);  
      printf("Test1.Byte01=%d \n",Test1.Byte01);  
    
      Data_Uint8_1=0x89;Data_Uint8_2=0xAB;Data_Uint8_3=0xCD;Data_Uint8_4=0xEF;
      printf("-------------------通过位运算合并-------------------\n");
      Data_Uint16_1 = (unsigned short int)(Data_Uint8_1 << 8) | (unsigned short int)(Data_Uint8_2);
      Data_Uint16_2 = (unsigned short int)(Data_Uint8_3 << 8) | (unsigned short int)(Data_Uint8_4);
      Data_Uint32=(unsigned int) (Data_Uint16_1 << 16) | (unsigned int)(Data_Uint16_2);
      printf("合并前的8位整数：0x%x,0x%x,0x%x,0x%x\n",Data_Uint8_1,Data_Uint8_2,Data_Uint8_3,Data_Uint8_4);
      printf("合并后的16位整数：0x%x,0x%x\n",Data_Uint16_1,Data_Uint16_2);
      printf("合并后的32位整数：0x%x\n",Data_Uint32);
    
    return 0;

\}  
输出：  
![C语言编程中的8位、16位、32位整数的分解与合并][C_8_16_32]

在这里练习以后进入Keil uVision5编程，想写个通用的转换函数，后面想没有必要，难道8、16位、32位相互转换这么复杂吗？如果这样，写在Keil uVision5里面的程序会是多么复杂，以后自己看起来也会繁琐的。

接着写：

直接强制转换：

printf("-------------------8位到32位转换-------------------\n");
      Data_Uint8_1=0x00;
      Data_Uint8_2=0x19;
      Data_Uint16_2=0x00;
      Data_Uint16_1 = (unsigned short int)(0x00 << 8) | (unsigned short int)(Data_Uint8_2);
      Data_Uint32=(unsigned int) (0x00 << 16) | (unsigned int)(Data_Uint16_1);
      printf("转换前的8位整数：0x%x,%d\n",Data_Uint8_2,sizeof(Data_Uint8_2));
      printf("8位到32位转换1======>转换后的32位整数：0x%x,数据长度：%d\n",Data_Uint32,sizeof(Data_Uint32));
      Data_Uint32=(unsigned int) (0x00 << 16) | (unsigned int)( (unsigned short int)(0x00 << 8) | (unsigned short int)(Data_Uint8_2));
      printf("8位到32位转换2======>转换后的32位整数：0x%x,数据长度：%d\n",Data_Uint32,sizeof(Data_Uint32));
      //Data_Uint32=(unsigned int) (0x00 << 24) | (unsigned int)(Data_Uint8_2);
      Data_Uint32= (unsigned int)(Data_Uint8_2);
      printf("8位到32位转换3======>转换后的32位整数：0x%x,数据长度：%d\n",Data_Uint32,sizeof(Data_Uint32));
      Data_Uint32=0x69;
      Data_Uint8_1= (unsigned char)(Data_Uint32);
      printf("32位整数：0x%x,数据长度：%d\n",Data_Uint32,sizeof(Data_Uint32));
      printf("32位到8位转换======>转换后的8位整数：0x%x,数据长度：%d\n",Data_Uint8_1,sizeof(Data_Uint8_1));

实际输出：

\-------------------8位到32位转换-------------------  
转换前的8位整数：0x19,1  
8位到32位转换1======>转换后的32位整数：0x19,数据长度：4  
8位到32位转换2======>转换后的32位整数：0x19,数据长度：4  
8位到32位转换3======>转换后的32位整数：0x19,数据长度：4  
32位整数：0x69,数据长度：4  
32位到8位转换======>转换后的8位整数：0x69,数据长度：1

说明：8位强制转32位，前面系统自动加了24位的0；32位强制转8位，系统只截取了最后的8位。

转载于:https://blog.51cto.com/dawn0919/2368268

[C_8_16_32]: https://s1.51cto.com/images/blog/201903/24/8215dfc3eead1a863047a2c117fe388f.jpg?x-oss-process=image/watermark,size_16,text_QDUxQ1RP5Y2a5a6i,color_FFFFFF,t_100,g_se,x_10,y_10,shadow_90,type_ZmFuZ3poZW5naGVpdGk=