我正在粘贴下面的客户端和服务器代码。我的程序运行正常，除了我试图在我的src和dest字段中发送一个ipaddress，由于某种原因，即使我发送它为131.199.166.232，它打印为232.166.199.131。但我的其余数据包值以适当的方式打印。我已经使用了memcpy（），所以我觉得它是一个memcpy的东西，在某个地方我做错了，但在Beej的指南中，因为有一个部分@在不同的计算机体系结构中的字节排序.....我没有使用过htonl（）所有，也许是因为那个。请指导我，因为我出错了。另外请告诉我发送数据的方式，如何在我的代码中使用htonl（）函数....提前感谢。
客户：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <netdb.h>
#define MAXPROFILES 2
int main(int argc, char *argv[])
{
int sockfd, portno, n;
struct sockaddr_in serv_addr;
struct hostent *server;
unsigned char buf[1024];
unsigned int srcAddress = 2193598184;
unsigned int destAddress = 2193598182;
struct profile_t
{
unsigned char length;
unsigned char type;
unsigned char *data;
};
typedef struct profile_datagram_t
{
unsigned char src[4];
unsigned char dst[4];
unsigned char ver;
unsigned char n;
struct profile_t profiles[MAXPROFILES];
} header;
header outObj;
int j =0;
int i =0;
// for loop for doing the malloc so that we can allocate memory to all profiles
for(i=0;i<MAXPROFILES;i++){
outObj.profiles[i].data = malloc(5);
}
for(i=3;i>=0;i--){
outObj.src[i] = (srcAddress >> (i*8)) & 0xFF;
outObj.dst[i] = (destAddress >> (i*8)) & 0xFF;
printf("%d",outObj.src[i]);
}
outObj.ver = 1;
outObj.n = 2;
memcpy(buf,&outObj.src,4);
memcpy(buf+4,&outObj.dst,4);
memcpy(buf+8,&outObj.ver,1);
memcpy(buf+9,&outObj.n,1);
outObj.profiles[0].length = 5;
outObj.profiles[0].type = 1;
outObj.profiles[1].length = 5;
outObj.profiles[1].type = 2;
for(i=0;i<MAXPROFILES;i++){
for(j=0;j<5;j++){
outObj.profiles[i].data[j] = j+1;
}
}
int k = 10;
// for loop to do memcopy of length,type and data.
for(i=0;i<MAXPROFILES;i++){
memcpy(buf+k,&outObj.profiles[0].length,1);
memcpy(buf+k+1,&outObj.profiles[0].type,1);
memcpy(buf+k+2,outObj.profiles[0].data,5);
k +=7;
}
if (argc < 3) {
fprintf(stderr,"usage: %s hostname port\n", argv[0]);
exit(0);
}
portno = atoi(argv[2]); //Convert ASCII to integer
sockfd = socket(AF_INET, SOCK_STREAM, 0); // socket file descriptor
if (sockfd < 0)
error("ERROR DETECTED !!! Problem in opening socket\n");
server = gethostbyname(argv[1]);
if (server == NULL) {
fprintf(stderr,"ERROR DETECTED !!!, no such server found \n");
exit(0);
}
bzero((char *) &serv_addr, sizeof(serv_addr)); //clear the memory for server address
serv_addr.sin_family = AF_INET;
bcopy((char *)server->h_addr,
(char *)&serv_addr.sin_addr.s_addr,
server->h_length);
serv_addr.sin_port = htons(portno);
printf("Client 1 trying to connect with server host %s on port %d\n", argv[1], portno);
if (connect(sockfd,(struct sockaddr *)&serv_addr,sizeof(serv_addr)) < 0)
error("ERROR in connection");
printf("SUCCESS !!! Connection established \n");
if (write(sockfd, buf, k) < 0)
{
error("Write error has occured ");
}
return 0;
服务器：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#define MAXPROFILES 2
int main(int argc, char *argv[])
{
int sockfd, newsockfd, portno, clilen;
struct sockaddr_in serv_addr, cli_addr;
unsigned char buf[1024];
int my_data2[10] = {1,3,9,10};
int my_data[10] = {1,2,3,4,5};
int myDataBinary[500] = {0};
int myDataBinary2[500] = {0};
int recData[500] = {0};
int index1=0;
struct profile_t
{
unsigned char length;
unsigned char type;
unsigned char *data;
};
typedef struct profile_datagram_t
{
unsigned char src[4];
unsigned char dst[4];
unsigned char ver;
unsigned char n;
struct profile_t profiles[MAXPROFILES];
} header;
header outObj;
int j =0;
int i =0;
if (argc < 2) {
fprintf(stderr,"usage: %s port_number1",argv[0]);
exit(1);
}
sockfd = socket(AF_INET, SOCK_STREAM, 0);
if (sockfd < 0)
error("ERROR DETECTED !!! Problem in opening socket");
bzero((char *) &serv_addr, sizeof(serv_addr));
portno = atoi(argv[1]);
serv_addr.sin_family = AF_INET;
serv_addr.sin_addr.s_addr = htonl(INADDR_ANY);
serv_addr.sin_port = htons(portno);
if (bind(sockfd, (struct sockaddr *) &serv_addr, sizeof(serv_addr)) < 0)
error("ERROR DETECTED !!! There was a problem in binding");
listen(sockfd, 10);
clilen = sizeof(cli_addr);
printf("Server listening on port number %d...\n", serv_addr.sin_port);
newsockfd = accept(sockfd,(struct sockaddr *) &cli_addr, &clilen);
if (newsockfd < 0)
error("ERROR DETECTED !!! the connection request was not accepted");
int rc = read(newsockfd,buf,100);
if(rc < 0){
printf("error");
}
else {
printf("success %d",rc);
}
memcpy(&outObj.src,buf+0,4);
memcpy(&outObj.dst,buf+4,4);
memcpy(&outObj.ver,buf+8,1);
memcpy(&outObj.n,buf+9,1);
printf("\nsrc ip = ");
for(int i=0;i<4;i++){
printf("%d ",outObj.src[i]);
}
printf("\ndest ip = ");
for(int i=0;i<4;i++){
printf("%d ",outObj.src[i]);
}
printf("\nversion = %d",outObj.ver);
printf("\nnumber = %d",outObj.n);
int k = 10;
for(i=0;i<outObj.n;i++){
memcpy(&outObj.profiles[i].length,buf+k,1);
memcpy(&outObj.profiles[i].type,buf+k+1,1);
outObj.profiles[i].data = malloc(outObj.profiles[i].length);
memcpy(outObj.profiles[i].data,buf+k+2,5);
k +=7;
}
for(int i=0;i<outObj.n;i++){
printf("\nMessage %d :",i+1);
printf("\nLength : %d",outObj.profiles[i].length);
printf("\nType : %d",outObj.profiles[i].type);
for(int j=0;j<5;j++){
printf("\ndata %d : %d",j,outObj.profiles[i].data[j]);
}
}
for(int i=0; i<sizeof(my_data)/sizeof(int);i++)
{
if(my_data[i] > 0){
index1 = my_data[i];
myDataBinary[index1] = 1;
printf("my data %d = %d\n",index1,myDataBinary[index1]);
}
}
for(int i=0; i<sizeof(my_data2)/sizeof(int);i++)
{
if(my_data2[i] > 0){
index1 = my_data2[i];
myDataBinary2[index1] = 1;
printf("my data %d = %d\n",index1,myDataBinary2[index1]);
}
}
int sumRecievedData = 0;
int sumMyData = 0;
int sumMultpliedData = 0;
float Cov =0;
float sdMyData = 0;
float sdRecievedData =0;
int n = 500;
float rho;
for(int i=0;i<outObj.n;i++){
index1=0;
for (int j=0; j<outObj.profiles[i].length;j++) {
if(outObj.profiles[i].data[j] > 0){
index1 = outObj.profiles[i].data[j];
recData[index1] = 1;
printf("rec data %d = %d\n",index1,recData[index1]);
}
}
}
return 0;
}
答案 0 :(得分：1)
ip地址实际上只是unsigned char的数组。
uchar ip[] = {127,0,0,1};
是环回地址的精确表示。但是一个包含四个字节的数组，而int实际上并不是那么大;有一个例外 endiannes ！所以假设我创建了代表该ip的int。一种天真的方法可能是：
int ip = (127<<24)|(0<<16)|(0<<8)|(1)
当然，在 little endian 计算机上，就像x86和arm一样：
char *char_ip = (void*)&ip;
并迭代该aray将产生：
1, 0, 0, 127
但是在 big endian 计算机上，如PowerPC或SPARC，我们会得到我们期望的结果，
127, 0, 0, 1
Big endian也被称为“网络字节顺序”，这是htonl中的n代表：“host to network long”。通过网络读取或写入整数时经常使用这些函数。假设服务器想要向客户端发送一些号码：
uint32_t important = htonl(42);
write(client, &important, sizeof important);
然后，要阅读它，客户端会：
uint32_t important;
read(server, &important, sizeof important);
important = ntohl(important);
您的IP地址被翻转的原因是因为IP地址预计将按网络字节顺序排列，但您的IP地址则为小端。 htonl上的int - 类型的ip会为你翻转它。
答案 1 :(得分：0)
这对我来说太多代码了，但是如果你只是想创建一个字节数组来进行序列化，你根本不需要任何特殊的函数，你可以用代数编写它：
unsigned char buf[sizeof(uint_type)];
uint_type value;
for (size_t i = 0; i != sizeof(uint_type); ++i)
{
// Little-endian
buf[i] = value >> (i * CHAR_BIT);
// Big-endian
buf[sizeof(uint_type - i - 1)] = value >> (i * CHAR_BIT);
}
请注意，此代码与系统的字节顺序无关。
自从你问到，htonl / ntohl函数是直接处理uint32_t整数，没有显式字符数组：
uint32_t value;
myfile.read((char*)&value, sizeof(value)); // network endianness
value = ntohl(value); // host endianness
/* fiddle fiddle */
value = htonl(value);
yourfile.write((const char*)&value, sizeof(value));
（我想，更常见的是你要写一个套接字而不是一个文件。）