Meaning of *& and **& in C++

Question

I found these symbols in a function declaration several times, but I don't know what they mean.

Example:

void raccogli_dati(double **& V, double **p, int N) { 
  int ultimo = 3; 
  V = new double * [N/2]; 
  for(int i=0; i < N/2; i++) { 
    V[i] = new double[N/2], std :: clog << "digita " << N/2 - i
                 << " valori per la parte superiore della matrice V: "; 
    for(int j=i; j < N/2; j++) 
      std :: cin >> V[i][j], p[ultimo++][0] = (V[i][j] /= sqrt(p[i][0]*p[j][0]));
  } 
  for(int i=1; i < N/2; i++) 
    for(int j=0; j < i; j++) 
       V[i][j] = V[j][i];
}

Answer 1

First is a reference to a pointer, second is a reference to a pointer to a pointer. See also FAQ on how pointers and references differ.

void foo(int*& x, int**& y) {
    // modifying x or y here will modify a or b in main
}

int main() {
    int val = 42;
    int *a  = &val;
    int **b = &a;

    foo(a, b);
    return 0;
}

Answer 2

That's passing a pointer by reference rather than by value. This for example allows altering the pointer (not the pointed-to object) in the function is such way that the calling code sees the change.

Compare:

void nochange( int* pointer ) //passed by value
{
   pointer++; // change will be discarded once function returns
}

void change( int*& pointer ) //passed by reference
{
   pointer++; // change will persist when function returns
}

Answer 3

int* is a pointer to an int. Then int*& must be a reference to a pointer to an int. Similarly, int** is a pointer to a pointer to an int, then int**& must be a reference to a pointer to a pointer to an int.

Answer 4

*& signifies the receiving the pointer by reference. It means it is an alias for the passing parameter. So, it affects the passing parameter.

#include <iostream>
using namespace std;

void foo(int *ptr)
{
    ptr = new int(50);    // Modifying the pointer to point to a different location
    cout << "In foo:\t" << *ptr << "\n"; 
    delete ptr ;
}

void bar(int *& ptr)
{
    ptr = new int(80);    // Modifying the pointer to point to a different location
    cout << "In bar:\t" << *ptr << "\n"; 
    // Deleting the pointer will result the actual passed parameter dangling
}
int main()
{
    int temp = 100 ;
    int *p = &temp ;

    cout << "Before foo:\t" << *p << "\n";
    foo(p) ;
    cout << "After foo:\t" << *p << "\n";

    cout << "Before bar:\t" << *p << "\n";
    bar(p) ;
    cout << "After bar:\t" << *p << "\n";

    delete p;

    return 0;
}

Output:

Before foo: 100
In foo: 50
After foo:  100
Before bar: 100
In bar: 80
After bar:  80

Answer 5

To understand those phrases let's look at the couple of things:

typedef double Foo;
void fooFunc(Foo &_bar){ ... }

So that's passing a double by reference.

typedef double* Foo;
void fooFunc(Foo &_bar){ ... }

now it's passing a pointer to a double by reference.

typedef double** Foo;
void fooFunc(Foo &_bar){ ... }

Finally, it's passing a pointer to a pointer to a double by reference. If you think in terms of typedefs like this you'll understand the proper ordering of the & and * plus what it means.

Answer 6

That is taking the parameter by reference. So in the first case you are taking a pointer parameter by reference so whatever modification you do to the value of the pointer is reflected outside the function. Second is the simlilar to first one with the only difference being that it is a double pointer. See this example:

void pass_by_value(int* p)
{
    //Allocate memory for int and store the address in p
    p = new int;
}

void pass_by_reference(int*& p)
{
    p = new int;
}

int main()
{
    int* p1 = NULL;
    int* p2 = NULL;

    pass_by_value(p1); //p1 will still be NULL after this call
    pass_by_reference(p2); //p2 's value is changed to point to the newly allocate memory

    return 0;
}