What does *& mean in a function parameter

Question

If I have a function that takes int *&, what does it means? How can I pass just an int or a pointer int to that function?

function(int *&am         


        

Answer 1

It's a reference to a pointer to an int. This means the function in question can modify the pointer as well as the int itself.

You can just pass a pointer in, the one complication being that the pointer needs to be an l-value, not just an r-value, so for example

int myint;
function(&myint);

alone isn't sufficient and neither would 0/NULL be allowable, Where as:

int myint;
int *myintptr = &myint;
function(myintptr);

would be acceptable. When the function returns it's quite possible that myintptr would no longer point to what it was initially pointing to.

int *myintptr = NULL;
function(myintptr);

might also make sense if the function was expecting to allocate the memory when given a NULL pointer. Check the documentation provided with the function (or read the source!) to see how the pointer is expected to be used.

Answer 2

This is a reference to a pointer to int - you would have to pass in the address of an int to this function, and be aware that the function could change the pointer through the reference.

Dumb example:

void func(int*& iref) 
{
    iref = new int;
}

int main()
{
    int i(0);
    int* pi(&i);

    func(pi);
    // pi no longer equal to &i

    return 0;
}

Answer 3

It means a reference to a pointer to an int. In other words, the function can change the parameter to point to something else.

To pass a variable in, just pass an int*. As awoodland points out, what's passed in must be an l-value.

Edit:

To build on awoodland's example:

#include <iostream>

void foo(int*& var)
{
    delete var;
    var = new int;
}

int main(int argc, char* argv[])
{
    int* var = NULL;

    std::cout << var << std::endl;

    foo(var);   // this function can/will change the value of the pointer

    std::cout << var << std::endl;

    delete var;

    return 0;
}

Answer 4

Simply: a reference to a pointer.

In C, without references, the traditional way to "relocate" a pointer, is to pass a pointer to a pointer:

void c_find(int** p, int val); /* *p will point to the node with value 'val' */

In C++, this can be expressed by the reference syntax, to avoid the awkward double dereference.

void cpp_find(int*& p, int val); // p will point to the node with value 'val'

Answer 5

function takes a single parameter, mynumber which is a reference to a pointer to an int.

This is useful when you need to pass a pointer to a function, and that function might change the pointer. For example, if you function is implemented like this:

function(int*& mynumber)
{
  if( !mynumber )
    mynumber = new int;
  *mynumber = 42;
}

...Then something like this might happen in the calling code:

int main()
{
  int* mynumber = 0;
  function(mynumber); // function will change what "mynumber" points to
  cout << *mynumber;
  return 0;
}