Is ->second defined for iterator my_map.end()?

Question

I'm working with a std::map<std::string, MyClass* >.

I want to test if my_map.find(key) returned a specific pointer.

Right now I'm doing;

auto iter = my_map.find(key);
if ((iter != my_map.end()) && (iter->second == expected)) {
    // Something wonderful has happened
}

However, the operator * of the iterator is required to return a reference. Intuitively I'm assuming it to be valid and fully initialized? If so, my_map.end()->second would be NULL, and (since NULL is never expected), I could reduce my if statement to:

if (iter->second == expected)

Is this valid according to specification? Does anyone have practical experience with the implementations of this? IMHO, the code becomes clearer, and possibly a tiny performance improvement could be achieved.

Answer 1

Intuitively I'm assuming it to be valid and fully initialized?

You cannot assume an iterator to an element past-the-end of a container to be dereferenceable. Per paragraph 24.2.1/5 of the C++11 Standard:

Just as a regular pointer to an array guarantees that there is a pointer value pointing past the last element of the array, so for any iterator type there is an iterator value that points past the last element of a corresponding sequence. These values are called past-the-end values. Values of an iterator i for which the expression *i is defined are called dereferenceable. The library never assumes that past-the-end values are dereferenceable. [...]