std::equal_range with lambda

Question

Lets say I have a vector of strings and I want to find all strings, that start with 'a', so I can do this:

struct cmp {
    bool operator()( const std::string &s, char c ) const { return s.front() < c; }
    bool operator()( char c, const std::string &s ) const { return s.front() < c; }
};
std::vector<std::string> strings;
...
std::sort( strings.begin(), strings.end() );
auto range = std::equal_range( strings.begin(), strings.end(), 'a', cmp{} );
...

This method is error prone, as it is easy to make mistake (for example I think it should be c < s.front() in second method) and has code duplication.

So is it possible to implement comparison function with generic lambda instead of structure with 2 methods?

More generic question, why value to compare has to be passed as argument to std::lower_bound, std::upper_bound and std::equal_range when it can easily be captured by lambda or passed to comparison structure, and then this issue will not be there at all?

How it could work if std::equal_range would not require value?

struct cmp {
    cmp( char lc ) : c( lc ) {}
    bool operator()( const std::string &s ) const { return s.front() < c; }
    char c;
};
std::vector<std::string> strings;
...
std::sort( strings.begin(), strings.end() );
auto range = std::equal_range( strings.begin(), strings.end(), cmp{'a'} );

Answer 1

lower_bound is

std::partition_point(strings.begin(), strings.end(),
                     [](const auto& s) { return s.front() < 'a'; });

upper_bound is

std::partition_point(strings.begin(), strings.end(),
                     [](const auto& s) { return s.front() <= 'a'; });

Yes, that means you have to write two calls to get the equivalent of equal_range. You can wrap this into a free template:

template<class Iter, class T, class Proj>
std::pair<Iter, Iter> my_equal_range(Iter first, Iter last, const T& value, Proj proj) {
    auto b = std::partition_point(first, last, [&](const auto& s) { return proj(s) < value; });
    auto e = std::partition_point(b, last, [&](const auto& s) { return !(value < proj(s)); });
    return {b, e};
}

And call it as

my_equal_range(strings.begin(), strings.end(), 'a',
               [](const auto& s) { return s.front(); });

---

The Ranges TS working draft adds projections to algorithms, so you will (eventually) be able to do this:

std::experimental::ranges::equal_range(strings, 'a', {},
                                       [](const auto& s) { return s.front(); });