default value of const reference parameter - c++

Question

Currently I have two functions:
void foo(const A & a) {
...
result = ...
result += handle(a); // 1
bar(result);
}
void foo() {
...
result = ...
bar(result);
}
All code in foo() are the same except 1.
Can I merge them to one function like following?
void foo(const A & a = 0) {
...
...
if (a) result += handle(a); // this won't work, but can I do something similar?
bar(result);
}
Btw, the parameter has to be a reference as I would like to keep the interface unchanged.

Answer 1

You can use the Null Object Pattern.
namespace
{
const A NULL_A; // (possibly "extern")
}
void foo(const A & a = NULL_A) {
...
result = ...
if (&a != &NULL_A) result += handle(a);
bar(result);
}

Answer 2

No. A reference is always an alias for a real object (assuming you don't trigger undefined behavior). You can achieve similar behavior without code duplication by accepting a pointer instead:
void foo_impl(A const* obj) {
// code as before, except access obj with -> rather than .
}
void foo (A const& obj) {
foo_impl(&obj);
}
void foo() {
foo_impl(nullptr);
}

Answer 3

In the spirit of DRY, why not merge them as such?
void foo(const A & a) {
foo();
handle(a);
}
void foo() {
...
...
}

Answer 4

The entire idea of using reference is that you avoid the NULL pointer issue. The reference nothing but an alias for the real object. I have an another simple idea for your program, basically you want to achieve two functionality with same function - using default parameters. Here is the code. Please excuse me for the variable names.
class ABC
{
public:
int t;
ABC operator=(ABC& other)
{
other.t = 0;
}
};
ABC other;
void foo( ABC &a=other);
void foo( ABC &a)
{
if( a.t == 0)
qDebug()<<"The A WAS ZERO";
else
qDebug()<<"THE A isn't zero";
}
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
ABC hello;
hello.t = 100;
foo();
foo(hello);
return a.exec();
}
Here is the output.
The A WAS ZERO
THE A isn't zero

Answer 5

Use a base class that provides an interface, and two derived classes that implement the interface, one using A and the other using nothing.
Refactor foo to use a common foo.
struct Handler
{
virtual int get() = 0;
};
struct AHandler : Handler
{
AHandler(const A& a) : a_(a) {}
virtual int get() { return handle(a_); }
const A& a_;
}
struct NullHandler : Handler
{
virtual int get() { return 0; }
}
void foo(const Handler & h) {
...
result = ...
result += h.get();
bar(result);
}
void foo(const A & a) {
AHandler ah(a);
foo(ah);
}
void foo() {
NullHandler nh(a);
foo(nh);
}