c - *** 检测到 glibc *** ./m : free(): invalid next size (fast) on trying to free heap memory

标签 c linux memory-management free glibc

<分区>

我有一个小代码,它在堆上分配一些内存,然后在使用后尝试释放它。 我在使用内存映射时遇到此错误。

代码是::

void merge(int *arr, int start, int mid, int end)
{
     int i = 0, j = 0, k = 0, tmp1 = 0, siz = 0;
     int *pt = NULL;
     j = start, k = mid+1, tmp1 = start;

     siz = (end - start) + 1;

     if ((pt = malloc(siz * sizeof(int))) == NULL) /* line 32 */
     {
         printf("\n ERROR Aloocating mem \n");
         return;
     }

     while (j <= mid && k <= end)
     {
        if (arr[j] <= arr[k])
        {
          pt[tmp1] = arr[j];
          ++j;
          ++tmp1;
        }
        if (arr[j] > arr[k])
        {
          pt[tmp1] = arr[k]; /* line 48 */
          ++k;
          ++tmp1;
        }

      } 
       while (j <= mid)
       {
          pt[tmp1] = arr[j]; /* line 56 */
          ++tmp1;
          ++j;
       }
        while (k <= end)
        {
           pt[tmp1] = arr[k];
           ++k;
           ++tmp1;
        }

        for (i = 0; i < siz; i++)
        {
           arr[end] = pt[end]; /* line 69 */
           --end;
        }   

        free(pt);   
    }   

我得到的错误是::

*** glibc detected *** ./m: free(): invalid next size (fast): 0x0940c008 ***
======= Backtrace: =========
/lib/tls/i686/cmov/libc.so.6(+0x6b161)[0x213161]
/lib/tls/i686/cmov/libc.so.6(+0x6c9b8)[0x2149b8]
/lib/tls/i686/cmov/libc.so.6(cfree+0x6d)[0x217a9d]
./m[0x80486d8]
./m[0x8048562]
./m[0x8048526]
./m[0x8048542]
./m[0x804874b]
/lib/tls/i686/cmov/libc.so.6(__libc_start_main+0xe6)[0x1bebd6]
./m[0x8048441]
======= Memory map: ========
001a8000-002fb000 r-xp 00000000 08:01 1535       /lib/tls/i686/cmov/libc-2.11.1.so
002fb000-002fd000 r--p 00153000 08:01 1535       /lib/tls/i686/cmov/libc-2.11.1.so
002fd000-002fe000 rw-p 00155000 08:01 1535       /lib/tls/i686/cmov/libc-2.11.1.so
002fe000-00301000 rw-p 00000000 00:00 0 
00483000-0049e000 r-xp 00000000 08:01 1712996    /lib/ld-2.11.1.so
0049e000-0049f000 r--p 0001a000 08:01 1712996    /lib/ld-2.11.1.so
0049f000-004a0000 rw-p 0001b000 08:01 1712996    /lib/ld-2.11.1.so
007f8000-00815000 r-xp 00000000 08:01 1708381    /lib/libgcc_s.so.1
00815000-00816000 r--p 0001c000 08:01 1708381    /lib/libgcc_s.so.1
00816000-00817000 rw-p 0001d000 08:01 1708381    /lib/libgcc_s.so.1
00b38000-00b39000 r-xp 00000000 00:00 0          [vdso]
08048000-08049000 r-xp 00000000 08:01 1317230    /home/abhi/Desktop/ad/A1/misc/m
08049000-0804a000 r--p 00000000 08:01 1317230    /home/abhi/Desktop/ad/A1/misc/m
0804a000-0804b000 rw-p 00001000 08:01 1317230    /home/abhi/Desktop/ad/A1/misc/m
0940c000-0942d000 rw-p 00000000 00:00 0          [heap]
b7600000-b7621000 rw-p 00000000 00:00 0 
b7621000-b7700000 ---p 00000000 00:00 0 
b777e000-b777f000 rw-p 00000000 00:00 0 
b778c000-b7790000 rw-p 00000000 00:00 0 
bf8b5000-bf8ca000 rw-p 00000000 00:00 0          [stack]
Aborted

查看了SO上一些类似问题的答案,得知是内存错误。

所以我用 valgrind 运行编码并得到以下输出(不是完整的输出)

==4494== 
==4494== Invalid write of size 4
==4494==    at 0x8048650: merge (merge.c:48)
==4494==    by 0x8048561: merge_sort (merge.c:20)
==4494==    by 0x8048525: merge_sort (merge.c:18)
==4494==    by 0x8048541: merge_sort (merge.c:19)
==4494==    by 0x8048766: main (merge.c:89)
==4494==  Address 0x41920ac is 4 bytes after a block of size 8 alloc'd
==4494==    at 0x4023AB8: malloc (vg_replace_malloc.c:207)
==4494==    by 0x80485CB: merge (merge.c:32)
==4494==    by 0x8048561: merge_sort (merge.c:20)
==4494==    by 0x8048525: merge_sort (merge.c:18)
==4494==    by 0x8048541: merge_sort (merge.c:19)
==4494==    by 0x8048766: main (merge.c:89)
==4494== 
==4494== Invalid write of size 4
==4494==    at 0x8048680: merge (merge.c:56)
==4494==    by 0x8048561: merge_sort (merge.c:20)
==4494==    by 0x8048525: merge_sort (merge.c:18)
==4494==    by 0x8048541: merge_sort (merge.c:19)
==4494==    by 0x8048766: main (merge.c:89)
==4494==  Address 0x41920b0 is 8 bytes after a block of size 8 alloc'd
==4494==    at 0x4023AB8: malloc (vg_replace_malloc.c:207)

==4494== Invalid read of size 4
==4494==    at 0x80486D5: merge (merge.c:69)
==4494==    by 0x8048561: merge_sort (merge.c:20)
==4494==    by 0x8048541: merge_sort (merge.c:19)
==4494==    by 0x8048766: main (merge.c:89)
==4494==  Address 0x41920ec is 8 bytes after a block of size 12 alloc'd
==4494==    at 0x4023AB8: malloc (vg_replace_malloc.c:207)
==4494==    by 0x80485CB: merge (merge.c:32)
==4494==    by 0x8048561: merge_sort (merge.c:20)
==4494==    by 0x8048541: merge_sort (merge.c:19)
==4494==    by 0x8048766: main (merge.c:89)

我无法纠正 malloc 对我来说看起来不错的问题。我正在分配 siz * sizeof(int) 字节(int 在我的机器上是 4 个字节),然后以数组形式使用分配的内存。

如果有人指出我正在做的错误并解释原因,将不胜感激。 谢谢

最佳答案

要扩展 Daniel 正确地指出 pt[end] 是不正确的,假设以下函数调用:

int values[] = { 0, 1, 2, 3, 4, 5, 6, 7 };
merge(values, 4, 5, 6);

merge 中,pt 被分配得足够大以容纳 (6 - 4) + 1 = 3 inttmp1 设置为 4。 在第 48 行,您执行 pt[tmp1] = arr[k],但这是 pt[4] = arr[0],并且您是在内存中写入在 pt 之外!

关于c - *** 检测到 glibc *** ./m : free(): invalid next size (fast) on trying to free heap memory,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14988217/

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