c - pcap_next() 将地址返回到不可访问的内存区域

Question

我正在编写一个 ARP 嗅探器，然后发生下一个运行时错误:

pcap_next() 函数返回指向地址空间的指针，该地址空间不可访问，仅在某些情况下会在运行时导致段错误。这是代码:

void function_arp(){
(...)

const unsigned char *frameRcv = NULL;
unsigned int byteNr = 0;
struct ether_arp *arpPack = NULL;

while (true) {
    // check on waiting time
    gettimeofday(&currentWaitTime, NULL);
    struct timeval diff = timediff(beginWaitTime, currentWaitTime);

    if (diff.tv_sec > 5) // wait for up to 5 seconds
    {
        fprintf(stderr, "Reading timed out\n");
        break;
    }

    // Receiving Frame
    byteNr = receiveRawFrame(handle, &frameRcv);
    if(frameRcv == 0x0) 
        continue;
    //when reading *frameRcv, the Segmentation fault occurs sometimes 
    fprintf(stdout, "%x - ", *frameRcv);  
    (...) 
    //The receiving of frames will be timed out after 4 sec
}

函数receiveRawFrame()的代码:

unsigned int
receiveRawFrame(struct capture_info handle, const unsigned char** receivedFrame)
{
    *receivedFrame = NULL;
    // try reading frames
    const unsigned char* frame;
    struct pcap_pkthdr pcapinfo;
    frame = pcap_next(handle.pcapHandle, &pcapinfo);

    *receivedFrame = frame;
    return pcapinfo.caplen;
}

一些注意事项:该程序将接收 IP 作为参数，并为该 IP 发送一个 ARP 请求，等待之后的回复。仅当 IP 不在本地网络中时才会发生 Segmentation Fault。在这种情况下，程序通常会超时。对于相同的输入(本地网络外的 IP)，有时会发生段错误，有时不会...

我假设在 while 循环中多次调用函数 pcap_next 时发生了意外情况。

操作系统:Debian 3.2.57-3 x86_64(64 位)Pcap 版本:1.15

GDB 提供下一个输出:

Program received signal SIGSEGV, Segmentation fault.
0x0000000000401b29 in function_arp () at src/arp_impl.c:120
120                             fprintf(stdout, "%x - ", *frameRcv);

Answer 1

检查 pcap_next 的结果。仅当结果非零时，才应访问框架和信息。