这篇文章主要讲解了“C语言编程中的8位、16位、32位整数的分解与合并方法是什么”,文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习“C语言编程中的8位、16位、32位整数的分解与合并方法是什么”吧!
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:
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。
感谢各位的阅读,以上就是“C语言编程中的8位、16位、32位整数的分解与合并方法是什么”的内容了,经过本文的学习后,相信大家对C语言编程中的8位、16位、32位整数的分解与合并方法是什么这一问题有了更深刻的体会,具体使用情况还需要大家实践验证。这里是亿速云,小编将为大家推送更多相关知识点的文章,欢迎关注!
免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。