段错误/使用大小为 8 的未初始化值答案

【问题标题】：Seg Fault / Use of uninitialised value of size 8段错误/使用大小为 8 的未初始化值
【发布时间】：2018-03-07 03:59:03
【问题描述】：

我知道 Seg 错误通常是由无效的内存访问引起的，所以我认为问题可能在于我如何初始化传递给 split 函数的二维动态数组。但是，我花了几个小时查看它，似乎找不到问题。

使用 valgrind 我得到以下信息：

==31912== Command: ./Project1 Data.txt
==31912== 
==31912== Use of uninitialised value of size 8
==31912==    at 0x4016C6: split(long**, long**, long**, long, long) (in /home/tony/Project1)
==31912==    by 0x400E5F: main (in /home/tony/Project1)
==31912== 
==31912== Invalid read of size 8
==31912==    at 0x4016C9: split(long**, long**, long**, long, long) (in /home/tony/Project1)
==31912==    by 0x400E5F: main (in /home/tony/Project1)
==31912==  Address 0x41d7894956415741 is not stack'd, malloc'd or (recently) free'd
==31912== 
==31912== 
==31912== Process terminating with default action of signal 11 (SIGSEGV)
==31912==  General Protection Fault
==31912==    at 0x4016C9: split(long**, long**, long**, long, long) (in /home/tony/Project1)
==31912==    by 0x400E5F: main (in /home/tony/Project1)
==31912== 
==31912== HEAP SUMMARY:
==31912==     in use at exit: 1,248 bytes in 81 blocks
==31912==   total heap usage: 89 allocs, 8 frees, 97,119 bytes allocated
==31912== 
==31912== LEAK SUMMARY:
==31912==    definitely lost: 304 bytes in 1 blocks
==31912==    indirectly lost: 304 bytes in 38 blocks
==31912==      possibly lost: 0 bytes in 0 blocks
==31912==    still reachable: 640 bytes in 42 blocks
==31912==         suppressed: 0 bytes in 0 blocks
==31912== Rerun with --leak-check=full to see details of leaked memory
==31912== 
==31912== For counts of detected and suppressed errors, rerun with: -v
==31912== Use --track-origins=yes to see where uninitialised values come from
==31912== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 0 from 0)
Segmentation fault (core dumped)

valgrind 似乎认为是罪魁祸首的 split 函数是：

void split(long** arr, long** arrL, long** arrR, long median, long size){   
//split data in half by x value -- takes in array sorted by x
    for (long i = 0; i < size -1; i++){
        if (i <= median){   //split data in half by x coord 
            arrL[i][0] = arr[i][0]; //x
            arrL[i][1] = arr[i][1]; //y
        }else{
            arrR[i][0] = arr[i][0];
            arrR[i][1] = arr[i][1];
        }
    }

}

在我的主函数中，我初始化数组，然后像这样调用 split()：

        arrL = new long*[median + 1];//add one to make sure there is enough room
        arrR = new long*[median + 1];
        for(long i = 0; i < median + 1; i++){
            arrL[i] = new long[1];
            arrR[i] = new long[1];
        }


    split(arr, arrL, arrR, median, size);   //splits data

任何帮助将不胜感激！

【问题讨论】：

如果您使用std::vector<std::vector<long>>，使用vector::at() 将或应该通过std::out_of_range 抛出异常向您显示错误。
如果您在启用调试信息的情况下进行编译 (gcc -g)，valgrind 通常会在报告问题时指向您的例程中的行号。你试过吗？
See this example。使用vector 然后at() 调用揭示了我的回答指出的错误。如果您仍然遇到问题，那么您很可能正在创建其他越界访问，可以使用该链接中的技术轻松诊断。

标签： c++ segmentation-fault valgrind dynamic-arrays

【解决方案1】：

仔细查看这条线（以及类似的线）。

arrL[i] = new long[1];

您分配了 1 个项目，因此 arrL[i] 的唯一有效索引是 0。但是在您的 split 函数中，您可以这样做：

arrL[i][1] = arr[i][1]; //y

这将访问索引1，超出范围。

您应该分配 2 个项目，而不是 1 个。

arrL[i] = new long[2];

【讨论】：

感谢您的回复。我解决了这个问题，但我仍然遇到同样的错误

【解决方案2】：

lMain 问题是您的 arrL 和 arrR 数组的大小为中位数，但写入 arrR 时 i 将大于中位数。

要么调整 arrL 和 arrR 的大小以包含 size *long 对象，要么根据以下代码为 arrR 使用 i-median 作为索引

void split(long** arr, long** arrL, long** arrR, long median, long size);

int _tmain(int argc, _TCHAR* argv[])
{
    int size = 10;
    int median = size / 2;

    long **arr, **arrL, **arrR;
    arr = new long*[size];
    for(long i = 0; i < size; i++){
        arr[i] = new long[2];
        arr[i][0] = i;
        arr[i][1] = i + 10;
    }

    arrL = new long*[median + 1];//add one to make sure there is enough room
    arrR = new long*[median + 1];
    for(int i = 0; i < median + 1; i++){
        arrL[i] = new long[2];
        arrR[i] = new long[2];
        arrL[i][0] = -1;
        arrL[i][1] = -1;
        arrR[i][0] = -1;
        arrR[i][1] = -1;
    }

    split(arr, arrL, arrR, median, size);

    for(int i = 0; i < median + 1; i++)
        printf("arrL %d = %ld %ld\n", i, arrL[i][0], arrL[i][1]);

    for(int i = 0; i < median + 1; i++)
        printf("arrR %d = %ld %ld\n", i, arrR[i][0], arrR[i][1]);

    return 0;
}

void split(long** arr, long** arrL, long** arrR, long median, long size){   
//split data in half by x value -- takes in array sorted by x
    for (int i = 0; i < size; i++){
        if (i < median){   //split data in half by x coord 
            arrL[i][0] = arr[i][0]; //x
            arrL[i][1] = arr[i][1]; //y
        }else{
            arrR[i - median][0] = arr[i][0];
            arrR[i - median][1] = arr[i][1];
        }
    }
}

【讨论】：