int的初始化与使用指针的字符串的初始化，c ++

Question

I recently ran into an issue while experimenting with linked lists. When I use a function to link up a new node that has a string in its data field it doesn't work. By this I mean that when the function (linkin() see below) returns , the string (which was local to the function) is destroyed and so the string field appears to be uninitialized.

However, when I do this exact same operation with an int it seems to work just fine. The code I used is below (its the int version, but make val a string instead of an int to see the other version). Could someone explain to me what is happening?

Thanks!

struct testlist {
    int val;
    testlist *next;
};
void linkin ( testlist *a );

int main() {

    testlist test;

    linkin(&test);
    cout << test.next->val <<endl;
}


void linkin ( testlist *a )
{
  testlist b;
  b.val=1;
  a->next = &b;
  cout << a->next->val <<endl;
}

Answer 1

testlist b;
a->next = &b;

a->next is pointing to a local temporary object which will be destroyed right after returning from function. It invokes undefined behavior after dereferencing it out of the function.

It's undefined behavior, sometimes it works, sometimes not.

---

Also in C++ there is a linked-list: std::list . On the other hand, you can use smart-pointers such as std::unique_ptr instead of bare pointers. I've written a smart-pointer based of your container:

struct List
{
    int val;
    unique_ptr<List> next;
};

void linkin (List &a)
{
    unique_ptr<List> b(new List);
    b->val = 1;
    a.next = move(b); // After move you can't use b anymore
    cout << a.next->val << endl;
}

int main()
{
    List test;
    linkin(test);
    cout << test.next->val <<endl;
}

Answer 2

Straight away, in linkin() I can see you're storing a pointer in a to an object that is going to go out of scope - ie: be destroyed.

There are reasons why this might appear to work with an int but not a complex type like a std::string. Despite appearances this is broken code - try rewriting linkin() :

void linkin ( testlist& a )
{
   testlist* b = new testlist;
   b->val=1;
   a->next = b;
   cout << a->next->val <<endl;
}