Hi folks,
My C++ knowledge is somewhat piecemeal. I was reworking some code at work. I changed a function to return a reference to a type. Inside, I look up an object based on an identifier passed in, then return a reference to the object if found. Of course I ran into the issue of what to return if I don't find the object, and in looking around the web, many people claim that returning a "null reference" in C++ is impossible. Based on this advice, I tried the trick of returning a success/fail boolean, and making the object reference an out parameter. However, I ran into the roadblock of needing to initialize the references I would pass as actual parameters, and of course there is no way to do this. I retreated to the usual approach of just returning a pointer.
I asked a colleague about it. He uses the following trick quite often, which is accepted by both a recent version of the Sun compiler and by gcc:
MyType& someFunc(int id)
{
// successful case here:
// ...
// fail case:
return *static_cast<MyType*>(0);
}
// Use:
...
MyType& mt = somefunc(myIdNum);
if (&mt) // test for "null reference"
{
// whatever
}
...
I have been maintaining this code base for a while, but I find that I don't have as much time to look up the small details about the language as I would like. I've been digging through my reference book but the answer to this one eludes me.
Now, I had a C++ course a few years ago, and therein we emphasized that in C++ everything is types, so I try to keep that in mind when thinking things through. Deconstructing the expression: "*static_cast(0);", it indeed seems to me that we take a literal zero, cast it to a pointer to MyType (which makes it a null pointer), and then apply the dereferencing operator in the context of assigning to a reference type (the return type), which should give me a reference to the same object pointed to by the pointer. This sure looks like returning a null reference to me.
Any advice in explaining why this works (or why it shouldn't) would be greatly appreciated.
Thanks,
Chuck
