What is the effect of & in c++ templates?

Question

template <typename T>
const T& longestOfTheThree(const T& a, const T& b, const T& c){
    T longest = a;
    if (longest.length < b.length){longest = b;}
    if (longest.length < c.length){longest = c;}
    return longest; 
}

Can someone explain why const T& (specifically the & before longestOfTheThree ) is invalid? Why does the code only compile if I remove the & ?

This code compiles:

template <typename T>
const T longestOfTheThree(const T& a, const T& b, const T& c){
    T longest = a;
    if (longest.length < b.length){longest = b;}
    if (longest.length < c.length){longest = c;}
    return longest; 
}

Answer 1

You are returning a reference to an ASDV ("stack" variable). That doesn't work: longest is destructed before return.

If you'd like to return a reference, you can do, for example:

template <typename T>
const T& longestOfTheThree(const T& a, const T& b, const T& c){
    if (longest.length < b.length) { return b; }
    if (longest.length < c.length) { return c; }
    return a;
}

Or, if you'd like to keep longest , you can have a pointer:

template <typename T>
const T& longestOfTheThree(const T& a, const T& b, const T& c){
    const T* longest = &a;
    if (longest->length < b.length) { longest = &b; }
    if (longest->length < c.length) { longest = &c; }
    return *longest;
}

Answer 2

Not sure why you try to implement your own template when there is std::max that will do almost what you need, but if you want to hide using of longest member I would do something like:

template<class T>
const T& longestOf(std::initializer_list<T> ilist)
{
    return *std::max_element(ilist.begin(), ilist.end(), [](const T& a, const T& b) {return a.longest < b.longest;});
}