typedef – non-optional in this case

Someone asked me to write a utility function to print any STL container, in my own loop. I suggested we follow the STL convention to use iterator inputs. Echoed on http://stackoverflow.com/questions/4657767/how-to-templatize-a-function-on-an-stl-container-that-has-an-iterator. However, what if we pass the container itself as input (assuming it’s a standard-conforming container)?

template
void dump(const CT& cont) {
    typedef typename CT::const_iterator iterator; //no choice
    iterator i;
    //won’t compile — CT::const_iterator i;
    for(i = cont.begin();   i!= cont.end();   ++i){
        cout<<*i <<" "<<endl;
    }
}

This works but the typedef isn’t sugar-coating. Without it you get

dependent-name ` M::const_iterator’ is parsed as a non-type, but instantiation yields a type

Very loosely, CT::const_iterator i suggests to the compiler to create a concrete type for i but CT::const_iterator is not a generic type, not “concretized” [1]. Solution — The typedef dresses up this “generic type” as if it’s an end-user type, usable in a variable declaration

[1] remember the Barcap FMD eval objects?

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