Argument passing by reference to pointer problem - c++

Question

Every time I try to compile my code I get error:
cannot convert parameter 1 from 'int *' to 'int *&'
The test code looks like this:
void set (int *&val){
*val = 10;
}
int main(){
int myVal;
int *pMyVal = new int;
set(&myVal); // <- this causes trouble
set(pMyVal); // <- however, this doesn't
}
I'd like to call that function in a single shot without creating a pointer somewhere only to pass it. And as pointers don't have constructors, something like this can't be done: set(int*(&myVal));
Is there any other way to pass a pointer by reference without needing to create a temporary variable?
Edit: By the way I know why the code fails to compile (I'm just passing the address which is possibly int and not an actual pointer). The question is how else can it be done.

Answer 1

A reference to non-const cannot bind to an rvalue. The result of the & operator is an rvalue. Take a look at the difference between lvalues and rvalues or read a good C++ book.
Also, in your context, you don't need to pass by reference. The following is OK as well:
void set (int *val){
*val = 10;
}
The reference would be needed if you were to do something like this;
void set (int*& val){
val = new int; //notice, you change the value of val, not *val
*val = 10;
}

Answer 2

&myval is an rvalue (of type int*), because it's a temporary. It's a pointer, but you cannot modify it, because it's just created on the fly. Your function set however requires a non-const reference, so you cannot pass it a temporary.
By contrast, pMyVal is a named variable, thus an lvalue, so it can be passed as a non-constant reference.

Answer 3

A very simple example can be found in this place.
http://markgodwin.blogspot.de/2009/08/c-reference-to-pointer.html

Answer 4

You can see the following sample code:
#include <iostream>
using namespace std;
void change(int*& ptr) {
cout << endl;
cout << "==================change(int*& ptr)====================" << endl;
cout << " &ptr = " << &ptr << endl;
cout << " ptr = " << ptr << endl;
cout << "=======================================================" << endl;
cout << endl;
*ptr *= *ptr;
}
int main(void) {
int* ptrNumber = new int(10);
cout << endl;
cout << "&ptrNumber = " << &ptrNumber << endl;
cout << "ptrNumber = " << ptrNumber << endl;
cout << ">>> *ptrNumber = " << *ptrNumber << endl;
change(ptrNumber);
cout << "<<< *ptrNumber = " << *ptrNumber << endl;
}
I installed Cygwin and used g++ to compile the above source code, binary file is out_pointer.exe.
Executing out_pointer.exe, output is as follows:
$ ./out_pointer.exe
&ptrNumber = 0x28ac3c
ptrNumber = 0x800102c0
>>> *ptrNumber = 10
==================change(int*& ptr)====================
&ptr = 0x28ac3c
ptr = 0x800102c0
=======================================================
<<< *ptrNumber = 100
From the above output, we see
&ptrNumber = &ptr
So, ptr is alias of ptrNumber. You can modify ptrNumber inside function void change(int*& ptr) by modifying ptr. For example, you can point ptr to another memory location as below:
#include <iostream>
using namespace std;
void change(int*& ptr) {
cout << endl;
cout << "==================change(int*& ptr)====================" << endl;
cout << " &ptr = " << &ptr << endl;
cout << " >>> ptr = " << ptr << endl;
ptr = new int(20);
cout << " <<< ptr = " << ptr << endl;
cout << "=======================================================" << endl;
cout << endl;
}
int main(void) {
int* ptrNumber = new int(10);
cout << endl;
cout << ">>> &ptrNumber = " << &ptrNumber << endl;
cout << ">>> ptrNumber = " << ptrNumber << endl;
cout << ">>> *ptrNumber = " << *ptrNumber << endl;
change(ptrNumber);
cout << "<<< &ptrNumber = " << &ptrNumber << endl;
cout << "<<< ptrNumber = " << ptrNumber << endl;
cout << "<<< *ptrNumber = " << *ptrNumber << endl;
}
New output:
$ ./out_pointer.exe
>>> &ptrNumber = 0x28ac3c
>>> ptrNumber = 0x800102c0
>>> *ptrNumber = 10
==================change(int*& ptr)====================
&ptr = 0x28ac3c
>>> ptr = 0x800102c0
<<< ptr = 0x80048328
=======================================================
<<< &ptrNumber = 0x28ac3c
<<< ptrNumber = 0x80048328
<<< *ptrNumber = 20

Answer 5

The problem is, int*&val can only be passed an lvalue, which the result of &myVal is not. By changing the signature to void set(int* const& val), it's telling the compiler you're not going to change the value of the pointer.
However, you normally wouldn't do that, only because if you're not going to change the value of the pointer, then passing the pointer by value is the most straightforward way to pass the value. And if you are going to change the value of the pointer, then you need to create a temporary to receive the result.