古代CPU一次至多能装载4字节(32位机),即一个整数，这4个字节在内存中的排列程序将影响它的值。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。
古代PC大多采纳小端字节序。当格式化的数据须要在两台应用不同字节序的主机进行传递时，接收端会谬误的解决。解决的办法是:发送端将发送的数据转换成大端字节序而后发送，接收端依据本人的字节序决定需不需要解决(小端转换，大端不解决)。
在做网络编程的时候，应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);如果平台没有提供这套，那么能够本人实现一下，原理很简略，就是调换下数据
#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下的测试后果如下图