How to ignore constness in a function template?

Question

I have the following template function:

template<typename K, typename V>
bool hasKey( const std::map<K, V>& m, K& k ) {
    return m.find(k) != m.end();
}

The keys in the map are not const.

Now, I may have a const K. How can I write a template that would allow me to pass in both K andconst K` to the function?

Is the solution to use a const_cast every time I call the function?

Maybe you'll get better results using a third template parameter for the item to search; then implicit conversions will be enabled from that parameter to `K` — M.M, Jul 03 '17 at 05:29
but if I pass in a const K, it still wouldnt match my map key type, would it? — chrise, Jul 03 '17 at 05:30

Curious · Accepted Answer · 2017-07-03T05:49:24.550

You can achieve what you want with the following

template <typename Key, typename Value, typename K>
bool hasKey(const std::map<Key, Value>& mp, const K& k) {
    return mp.find(k) != mp.end();
}

This way you are sure when looking at the declaration of the function that neither operand is going to be modified since they are both references to const.

Both non-const and const references (and even rvalues) to key types can be passed to the .find() methods in std::map. This works because the .find() method accepts a key type by const reference, so if you pass a non-const reference it gets bound by a const reference anyway, so it doesn't make a difference.

Another thing that you get from templating the key types separately is compatibility with transparent comparators (since C++14, see http://en.cppreference.com/w/cpp/container/map/find). For more on what transparent comparators are see What are transparent comparators?.

score 3 · Answer 2 · answered Jul 03 '17 at 05:56

3

You could have hasKey based on a concept of Map:

template<typename Map>
bool hasKey(const Map& m, const typename Map::key_type& k) {
    return m.find(k) != m.end();
}

answered Jul 03 '17 at 05:56

Cosme

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How to ignore constness in a function template?

2 Answers2