C ++使用指针

Question

I don't understand why it outputs "123" even though the deletenode function set x to NULL.

#include <iostream>;
using namespace std;
struct node {int value; struct node *left,*right,*parent;};
void deletenode(node *a) {
    delete a;
    a=NULL;
}
int main(int argc, const char * argv[])
{
    node *x = new node;
    x->value=123;
    deletenode(x);
    if (x!=NULL) cout << x->value;
}

Answer 1

You have the following signature: void deletenode(node *a) . You pass a pointer x to the function. The pointer value gets copied . Inside the function, you modify the local pointer value by writing to it via a = NULL .

However, that modification happens on the copy. The original remains unaffected. Notice that this isn't true for delete , since delete doesn't modify the pointer, it modifies (or rather, purges) the pointee .

The superficial solution is to pass the pointer value by reference:

void deletenode(node*& a)

However, there's a consensus that setting pointers to nullptr (or NULL ) after deletion doesn't really help, and is therefore not normally done. I would therefore replace your whole call to deletenode with a simple delete x; statement.

Answer 2

Although you've successfully deleted the object you intended to, the pointer itself is passed by value (copied to variable 'a') to deletenode(). So even though variable 'a' inside the function is null, variable 'x' is still pointing to the now-deleted memory.

Answer 3

因为使用delete删除实际上是将内存标记为空闲并可以重用，但是该区域的内容可能保持不变

Answer 4

You pass this pointer to the function BY VALUE - this means you wont see changes outside of the function. Try to pass it by reference.

Answer 5

#include <iostream>

template<typename T>
void destroy(T*& ptr) {
  delete ptr;
  ptr = NULL;
}

template<typename T>
struct node {
  node(T t) : value(t) { }
  ~node() {
    delete left;
    delete right;
  }
  T value;
  struct node *left,*right,*parent;
};

int main(int argc, const char * argv[]) {
  node<int> *x = new node<int>(123);

  destroy(x);

  if (x != NULL) {
    std::cout << x->value << '\n';
  }
}