使用malloc释放c中分配的空间

Question

我对此代码有疑问:

typedef struct pop {
unsigned long int *np; // matrix
unsigned long int f;
long double fp; 
unsigned long int *R; // matrix
unsigned long int *C; // matrix
unsigned long int Dp;
unsigned long int Ds;
unsigned long int count;
struct popolazione *ptrTempLst; // pointer
struct popolazione *leftTree;  // left tree pointer
struct popolazione *rightTree; // right tree pointer
} Node; 

当我释放为这个结构分配的空间时,之前我是否要在struct中释放指向矩阵的指针？

例如,

 Node *ptr=(Node *) malloc(sizeOf(Node));
 ptr->np=(unsigned long int *)malloc(10*sizeOf(unsigned long int));

 /*code code code*/

 // is necessary: free(ptr->np); 

 free(ptr);

提前致谢

Answer 1

是.

每次通话都malloc必须有匹配的电话free.

Answer 2

那是正确的.

为了避免自己在脚下射击,您可以考虑以下做法:

1. 始终释放所有malloc/calloc'ed内存与free()
2. 然后,将指针设置为NULL
3. 使用专用的清理/销毁功能来确保一致的内存清理

以下函数是确保始终正确清理结构,避免内存泄漏,并避免意外释放已释放的内存并导致分段错误的好方法:

int destroyNode(Node* myNode) {
  if(!myNode) {
    printf("Invalid pointer! Exiting");
    return (-1);
  }

  // Clear out memory
  if(np) {
    free(np);
    np = NULL;
  }
  if(R) {
    free(R);
    R = NULL;
  }
  if(C) {
    free(C);
    C = NULL;
  }
  if(ptrTempLst) {
    free(ptrTempLst);
    ptrTempLst = NULL;
  }
  if(leftTree) {
    free(leftTree);
    leftTree = NULL;
  }
  if(rightTree) {
    free(rightTree);
    rightTree = NULL;
  }

  free(myNode);
}

例如:

int main(void) {
  Node *tempNode = calloc((size_t)1,sizeof(Node));

  // Alloc the member nodes, etc, do some code


  // Ready to clean up and exit program
  destroyNode(tempNode);
  tempNode = NULL;

  return 0;
}

祝好运!