C++，获取网站源，第2部分

Question

所以，我有这个源：现在

#include <winsock2.h> 
#include <windows.h> 
#include <iostream> 
#pragma comment(lib,"ws2_32.lib") 

using namespace std; 

int main(){ 
     WSADATA wsaData; 

    if (WSAStartup(MAKEWORD(2,2), &wsaData) != 0) { 
       cout << "WSAStartup failed.\n"; 
     system("pause"); 
       return 1; 
    } 

     SOCKET Socket=socket(AF_INET,SOCK_STREAM,IPPROTO_TCP); 

     struct hostent *host; 
     host = gethostbyname("www.newegg.com"); 

     SOCKADDR_IN SockAddr; 
     SockAddr.sin_port=htons(80); 
     SockAddr.sin_family=AF_INET; 
     SockAddr.sin_addr.s_addr = *((unsigned long*)host->h_addr); 

     cout << "Connecting...\n"; 
     if(connect(Socket,(SOCKADDR*)(&SockAddr),sizeof(SockAddr)) != 0){ 
       cout << "Could not connect"; 
       system("pause"); 
       return 1; 
     } 
     cout << "Connected.\n"; 

     send(Socket,"GET/HTTP/1.1\r\nHost: www.newegg.com\r\nConnection: close\r\n\r\n", strlen("GET/HTTP/1.1\r\nHost: www.newegg.com\r\nConnection: close\r\n\r\n"),0); 
     char buffer[10000]; 

     int nDataLength; 
     while ((nDataLength = recv(Socket,buffer,10000,0)) > 0){    
       int i = 0; 
       while (buffer[i] || buffer[i] == '\n' || buffer[i] == '\r') { 
         cout << buffer[i]; 
         i += 1; 
       } 
     } 

     closesocket(Socket); 
     WSACleanup(); 

     system("pause"); 
     return 0; 
} 

，它的工作原理获取来源，然而，事后不断重复。例如：

http://img263.imageshack.us/img263/7259/outputq.png

注意它是如何说，然后继续？我想知道如何避免这种情况。我知道我可以限制缓冲区，但是有没有更好的解决方案？由于

Answer 1

尝试改变

while (buffer[i] || buffer[i] == '\n' || buffer[i] == '\r') { 
      cout << buffer[i]; 
      i += 1; 
    } 

到

for(; i != nDataLength; ++i) 
      cout << buffer[i]; 

它看起来像你正在阅读了缓冲区到由recv和朋友使用的垃圾数据的结束。没有人说，返回的数据必须是空终止的，这是while循环为了正确的行为而假定的。