C ++静态空指针

Question

I can't figure out the following situation: I've 3 classes:

AA.h

class AA
{
   public:
   AA();
   static CC *i;
};

BB.h

class BB
{
  public:
  BB();
  void static setI(CC *i);
};

CC.h

class CC
{
  public:
  CC();
};

AA.cpp

AA::AA(){}
CC *AA::i= nullptr;

BB.cpp

BB::BB(){}

void BB::setI(CC *i)
{
  i = new CC();
  cout<<i<<endl;
  cout<<AA::i;
}

CC.cpp

CC::CC(){}

So I've a static pointe of type CC in the class A. And i start the main in the following way:

int main(int argc, char *argv[])
{
    BB::setI(AA::i);
    return 0;
}

The output I get is

//0xaeed70
//0

So why AA::i does not equal i ?

Answer 1

The parameter type of BB::setI is CC *i , the pointer itself is passed by value, i is copied from the argument. Any modification on i inside the function has nothing to do with the original argument AA::i .

What you want might be pass-by-reference, ie

void BB::setI(CC *&i)
{
  i = new CC();
  cout<<i<<endl;
  cout<<AA::i;
}

Answer 2

In the expression i = new CC() variable i is local variable of function BB::setI which was initialized with the value of AA::i because you called function like BB::setI(AA::i); . Assigning any value to i will only change the value of this local variable, not the value of some other variable that has been used to initialize it when you called the function.

Answer 3

It's exactly the same situation as this smaller example:

int i = 0;

void setI(int x)
{
    x = 1234;
    cout << x << endl;
    cout << i;
}

int main()
{
    setI(i);
}

As you can see more clearly here, you're assigning a new value to the function's parameter.
This parameter is unrelated to AA 's member.

Regardless of its type, if you want a function to modify an object, you need to pass it a reference (or pointer) to that object; you can't pass the object's value.