将指针传递给函数不会返回值

Question

In following case I get NumRecPrinted = 0 , that is num is 0

int main()
{
    int demo(int *NumRecPrinted);
    int num = 0;
    demo(&num);
    cout << "NumRecPrinted=" << num;    <<<< Prints 0
    return 0;
}

int demo (int *NumRecPrinted)

{
    int no_of_records = 11;
    NumRecPrinted = &no_of_records;
}

Answer 1

No!

*NumRecPrinted = no_of_records;

See "*" means "value of" and "&" means "address of". You want to change the "value of" NumRecPrinted, which is why the above works. What you did was to give NumRecPrinted the "address of" num_of_records.

Answer 2

You are assigning the address to the pointer, and not the value to the pointed-to. Try it like this instead

int demo (int *NumRecPrinted)
{
     int no_of_records = 11;
     *NumRecPrinted = no_of_records; 
} 

Answer 3

All you did was pointer the local pointer-to-int NumRecPrinted at a new integer inside the demo function.

You want to change the integer it points to, not change where it points.

*NumRecPrinted = no_of_records;

You can see in your version that you're taking the address of a local variable, and you know it isn't the address of that variable you care about, but its value.

Answer 4

As others have pointed out, the * = the value of and & = address of. So you were just assigning a new address to the pointer inside the method. You should:

*NumRecPrinted = no_of_records; 

See this excellent tutorial on Pointers . Eg:

  int firstvalue = 5, secondvalue = 15;
  int * p1, * p2;

  p1 = &firstvalue;  // p1 = address of firstvalue
  p2 = &secondvalue; // p2 = address of secondvalue
  *p1 = 10;          // value pointed by p1 = 10
  *p2 = *p1;         // value pointed by p2 = value pointed by p1
  p1 = p2;           // p1 = p2 (value of pointer is copied)
  *p1 = 20;          // value pointed by p1 = 20

Answer 5

You want *NumRecPrinted = no_of_records;

That means, "set the thing NumRecPrinted points to to equal no_of_records".