C++实现主机字节序和网络字节序转换示例
作者：点墨
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字节序问题
现代CPU一次至少能装载4字节(32位机),即一个整数，这4个字节在内存中的排列顺序将影响它的值。这就是字节序问题。字节序分为大端字节序(big endian)和小端字节序(little endian)。
如上图，大端字节序是指数据的高位存在内存的低地址处，小端字节序是指数据的高位存在内存的高地址处。
现代PC大多采用小端字节序
当格式化的数据需要在两台使用不同字节序的主机进行传递时，接收端会错误的处理。解决的方法是:发送端将发送的数据转换成大端字节序然后发送，接收端根据自己的字节序决定需不需要处理(小端转换，大端不处理)。
在做网络编程的时候，使用提供的库即可进行转换(linux库,windows下是一样的，头文件有所区别)
#include <netinet/in.h> unsigned long int _htonl(unsigned long int hostlong); unsigned short int _htons(unsigned short int hostshort); unsigned long int _ntohl(unsigned long int netlong); unsigned short int _ntohs(unsigned short int netshort);
互换数据
如果平台没有提供这套，那么可以自己实现一下，原理很简单，就是互换下数据
#include <stdint.h>
#define BigLittleSwap16(v) ((0xFF00 & (uint16_t)(v)) >> 8 | (0x00FF & (uint16_t)(v)) << 8)
#define BigLittleSwap32(v) ((0xFF000000 & (uint32_t)(v)) >> 24 | (0x00FF0000 & (uint32_t)(v)) >> 8 | (0x0000FF00 & (uint32_t)(v)) << 8 | (0x000000FF & (uint32_t)(v)) << 24)
#define BigLittleSwap64(v) ((0xFF00000000000000 & (uint64_t)(v)) >> 56 |(0x00FF000000000000 & (uint64_t)(v)) >> 40 |(0x0000FF0000000000 & (uint64_t)(v) )>> 24 |(0x000000FF00000000 & (uint64_t)(v)) >> 8 |(0x00000000FF000000 & (uint64_t)(v)) << 8| (0x0000000000FF0000 & (uint64_t)(v)) << 24 | (0x000000000000FF00 & (uint64_t)(v) )<< 40 | (0x00000000000000FF & (uint64_t)(v) )<< 56 )
int isCPUBigEndian(){
union Data{
short val;
char byte;
} data;
data.val = 0x0102;
return (data.byte == 0x01);
}
unsigned long int _htonl(unsigned long int hostlong);
unsigned short int _htons(unsigned short int hostshort);
unsigned long int _ntohl(unsigned long int netlong);
unsigned short int _ntohs(unsigned short int netshort);
unsigned long int _htonl(unsigned long int hostlong){
return isCPUBigEndian() ? hostlong : BigLittleSwap32(hostlong);
}
unsigned short int _htons(unsigned short int hostshort){
return isCPUBigEndian() ? hostshort : BigLittleSwap16(hostshort);
}
unsigned long int _ntohl(unsigned long int netlong){
return isCPUBigEndian() ? netlong : BigLittleSwap32(netlong);
}
unsigned short int _ntohs(unsigned short int netshort){
return isCPUBigEndian() ? netshort : BigLittleSwap16(netshort);
}测试
#include <netinet/in.h>
#include <stdio.h>
int main(int argc,const char*argv[]){
long int v1 = 0x11223344;
short int v2 = 0x1122;
printf("the endian of system is:%s\n",isCPUBigEndian()?"Big Endian":"Little Endian");
printf("htonl:%x , %x , %x\n",v1,htonl(v1),_htonl(v1));
printf("htons:%x , %x , %x\n",v2,htons(v2),_htons(v2));
printf("ntohl:%x , %x , %x\n",v1,ntohl(v1),_ntohl(v1));
printf("ntohs:%x , %x , %x\n",v2,htons(v2),_htons(v2));
return 0;
}在linux下的测试结果如下图
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