Valgrind 坚持认为堆上还有内存，但我正在释放指向我拥有的堆的唯一指针

Question

我正在使用 Valgrind 查找我的代码中的任何内存泄漏，根据输出，仍有 17 个字节可以访问，1 个分配和 0 个释放：

==9413== 
==9413== HEAP SUMMARY:
==9413==     in use at exit: 17 bytes in 1 blocks
==9413==   total heap usage: 1 allocs, 0 frees, 17 bytes allocated
==9413== 
==9413== 17 bytes in 1 blocks are still reachable in loss record 1 of 1
==9413==    at 0x402C109: calloc (in /usr/lib/valgrind/vgpreload_memcheck-x86-linux.so)
==9413==    by 0x80485F6: main (question3.c:21)
==9413== 
==9413== LEAK SUMMARY:
==9413==    definitely lost: 0 bytes in 0 blocks
==9413==    indirectly lost: 0 bytes in 0 blocks
==9413==      possibly lost: 0 bytes in 0 blocks
==9413==    still reachable: 17 bytes in 1 blocks
==9413==         suppressed: 0 bytes in 0 blocks
==9413== 
==9413== For counts of detected and suppressed errors, rerun with: -v
==9413== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

问题是我的代码中只有一个 calloc，我肯定会释放指针：

void checkInput(int err) {
  if (!err || err == EOF) {
    printf("\nInvalid input!\n");
    exit(1);
  }
}

#define N 17
int main() {
  char bufferStack[N] = { 'a' };
  char * bufferHeap = calloc(N, sizeof(char));
  char junk[N];
  printf("Enter a string no more than 16 chars long\n0123456789012345\n");
  checkInput(scanf("%16[^\n]", bufferStack));
  printf("BufferStack: %s\n", bufferStack);
  checkInput(scanf("%16[\n]", junk));
  checkInput(scanf("%16[^\n]", bufferHeap));
  printf("BufferHeap: %s\n", bufferHeap);
  free(bufferHeap);
  return 0;
}

我不确定内存在何处（或如何）泄漏。如果有人可以帮助我找到我的错误或解决问题，我将不胜感激。谢谢

Answer 1

如果您在第一次提示时输入超过 16 个字符，第一次调用 scanf 将选取前 16 个字符，将所有剩余字符留在输入缓冲区中。

然后当第二个 scanf 执行时，它需要一个换行符。下一个字符不是换行符，而是第一个提示符留下的字符。这导致没有任何匹配，所以 scanf returns 0。这导致 checkInput 打印错误消息并调用 exit，它立即退出程序而不释放内存。这就是 "still reachable" 内存的来源。

您需要重组代码，以便所有代码路径 free 分配内存。

Valgrind 坚持认为堆上还有内存，但我正在释放指向我拥有的堆的唯一指针

Valgrind insists there's still memory on the heap but I'm freeing the only pointer to the heap I have

c

free

valgrind

calloc