C++，獲取網站源，第2部分

Question

所以，我有這個源：現在

#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; 
} 

，它的工作原理獲取來源，然而，事後不斷重複。例如：

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

注意它是如何說，然後繼續？我想知道如何避免這種情況。我知道我可以限制緩衝區，但是有沒有更好的解決方案？由於

Answer 1

嘗試改變

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

到

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

它看起來像你正在閱讀了緩衝區到由recv和朋友使用的垃圾數據的結束。沒有人說，返回的數據必須是空終止的，這是while循環爲了正確的行爲而假定的。