Returning a “NULL reference” in C++?

Question

In dynamically typed languages like JavaScript or PHP, I often do functions such as:

function getSomething(name) {
    if (content_[name]) return content_[na         


        

Answer 1

You cannot do this during references, as they should never be NULL. There are basically three options, one using a pointer, the others using value semantics.

1. With a pointer (note: this requires that the resource doesn't get destructed while the caller has a pointer to it; also make sure the caller knows it doesn't need to delete the object):

   SomeResource* SomeClass::getSomething(std::string name) {
       std::map::iterator it = content_.find(name);
       if (it != content_.end()) 
           return &(*it);  
       return NULL;  
   }
   

2. Using std::pair with a bool to indicate if the item is valid or not (note: requires that SomeResource has an appropriate default constructor and is not expensive to construct):

   std::pair SomeClass::getSomething(std::string name) {
       std::map::iterator it = content_.find(name);
       if (it != content_.end()) 
           return std::make_pair(*it, true);  
       return std::make_pair(SomeResource(), false);  
   }
   

3. Using boost::optional:

   boost::optional SomeClass::getSomething(std::string name) {
       std::map::iterator it = content_.find(name);
       if (it != content_.end()) 
           return *it;  
       return boost::optional();  
   }
   

If you want value semantics and have the ability to use Boost, I'd recommend option three. The primary advantage of boost::optional over std::pair is that an unitialized boost::optional value doesn't construct the type its encapsulating. This means it works for types that have no default constructor and saves time/memory for types with a non-trivial default constructor.

I also modified your example so you're not searching the map twice (by reusing the iterator).