如果没有找到搜索结果，返回一个“ NULL”对象

If you want a NULL return value you need to use pointers instead of references.

References can't themselves be NULL.

(Note to the future comment posters: Yes you can have the address of a reference be NULL if you really really try to).

See my answer here for a list of differences between references and pointers.

In C++, references can't be null. If you want to optionally return null if nothing is found, you need to return a pointer, not a reference:

Attr *getAttribute(const string& attribute_name) const {
//search collection
//if found at i
return &attributes[i];
//if not found
return nullptr;
}

Otherwise, if you insist on returning by reference, then you should throw an exception if the attribute isn't found.

(By the way, I'm a little worried about your method being const and returning a non-const attribute. For philosophical reasons, I'd suggest returning const Attr *. If you also may want to modify this attribute, you can overload with a non-const method returning a non-const attribute as well.)

You are unable to return NULL because the return type of the function is an object reference and not a pointer.

The reason that you can't return NULL here is because you've declared your return type as Attr&. The trailing & makes the return value a "reference", which is basically a guaranteed-not-to-be-null pointer to an existing object. If you want to be able to return null, change Attr& to Attr*.

There are several possible answers here. You want to return something that might exist. Here are some options, ranging from my least preferred to most preferred:

• Return by reference, and signal can-not-find by exception.

  Attr& getAttribute(const string& attribute_name) const
  {
  //search collection
  //if found at i
  return attributes[i];
  //if not found
  throw no_such_attribute_error;
  }

It's likely that not finding attributes is a normal part of execution, and hence not very exceptional. The handling for this would be noisy. A null value cannot be returned because it's undefined behaviour to have null references.

• Return by pointer

  Attr* getAttribute(const string& attribute_name) const
  {
  //search collection
  //if found at i
  return &attributes[i];
  //if not found
  return nullptr;
  }

It's easy to forget to check whether a result from getAttribute would be a non-NULL pointer, and is an easy source of bugs.

• Use Boost.Optional

  boost::optional<Attr&> getAttribute(const string& attribute_name) const
  {
  //search collection
  //if found at i
  return attributes[i];
  //if not found
  return boost::optional<Attr&>();
  }

A boost::optional signifies exactly what is going on here, and has easy methods for inspecting whether such an attribute was found.

Side note: std::optional was recently voted into C++17, so this will be a "standard" thing in the near future.

You can easily create a static object that represents a NULL return.

class Attr;
extern Attr AttrNull;


class Node {
....


Attr& getAttribute(const string& attribute_name) const {
//search collection
//if found at i
return attributes[i];
//if not found
return AttrNull;
}


bool IsNull(const Attr& test) const {
return &test == &AttrNull;
}


private:
vector<Attr> attributes;
};

And somewhere in a source file:

static Attr AttrNull;

You can try this:

return &Type();

As you have figured out that you cannot do it the way you have done in Java (or C#). Here is another suggestion, you could pass in the reference of the object as an argument and return bool value. If the result is found in your collection, you could assign it to the reference being passed and return ‘true’, otherwise return ‘false’. Please consider this code.

typedef std::map<string, Operator> OPERATORS_MAP;


bool OperatorList::tryGetOperator(string token, Operator& op)
{
bool val = false;


OPERATORS_MAP::iterator it = m_operators.find(token);
if (it != m_operators.end())
{
op = it->second;
val = true;
}
return val;
}

The function above has to find the Operator against the key 'token', if it finds the one it returns true and assign the value to parameter Operator& op.

The caller code for this routine looks like this

Operator opr;
if (OperatorList::tryGetOperator(strOperator, opr))
{
//Do something here if true is returned.
}

There is one more option that could be considered in this situation - depending on your design. You can return the value using an argument to your function and make the function return bool, e.g.

bool getAttribute(const string& attribute_name, Attr& returnAttr) const {
//search collection
//if found at i
returnAttr = attributes[i];
return true;
//if not found
return false;
}