c++expression-templatesrange-v3

Can algorithms be made compatible with expression templates?


Suppose I have some array-based code that is enabled to be used by expression templates. E.g., I have operator[] for these arrays overloaded and also have overloaded the arithmetic operator + etc.

Now I would like to let the STL algorithm any_of run on such arrays. The simple way is to do

ExprArray<double, N> b, c; // init etc. 
auto a = b + c;            // for (auto i = 0; i < N; ++i) { a[i] = b[i] + c[i]; }
auto res = std::any_of(begin(a), end(a), SomePred{});

Of course, I would like to be able to short-circuit the computation and have a modified (range-based) lib::any_of that does

// only compute b[i] + c[i] until SomePred is satisified
auto res = lib::any_of(b + c, SomePred{}); // write as explicit loop over b[i] + c[i]

Writing lib::any_of in terms of operator[] on its input will do that job, the same as it was done for the overloaded operator+. However, this would require similar reimplementations of all STL algorithms that I could possibly run on such arrays.

Question: So suppose I want to re-use existing range-based algorithms (Boost.Range, range-v3) by only modifying the ExprArray iterators. Is it possible to modify the ExprArray iterator operator* and operator++ in such a way that this is transparent to range-based algorithms?

// only compute b[i] + c[i] until SomePred is satisified
// needs to eventually dispatch to 
// for (auto i = 0; i < N; ++i)
//     if (SomePred(b[i] + c[i])) return true;
// return false;
auto res = ranges::any_of(b + c, SomePred{});

So if the algorithm version actually is implemented in terms of iterators, the loop for (auto it = first; it != last; ++it) needs *it to be aware of the fact that it needs to compute b[i] + c[i], and ++it has to know that it needs to do ++i.


Solution

  • This question seems to reduce to "Can I implement iterators for my expression templates?" which I think is fairly straightforward. Assuming that "expression templates" know their size and have overloaded operator[] an iterator simply needs to hold a reference to the expression object and an offset into the range it represents:

    template <class Expr>
    class iterator {
    public:
      using iterator_category = ranges::random_access_iterator_tag;
      using difference_type = std::ptrdiff_t;
      using value_type = typename Expr::value_type;
    
      iterator() = default;
      constexpr iterator(Expr& e, difference_type i) :
        expr_{&e}, i_{i} {}
    
      constexpr bool operator==(const iterator& that) const {
        return assert(expr_ == that.expr_), i_ == that.i_;
      }
      constexpr bool operator!=(const iterator& that) const {
        return !(*this == that);
      }
      // Similarly for operators <, >, <=, >=
    
      value_type operator*() const {
        return (*expr_)[i_];
      }
      value_type operator[](difference_type n) const {
        return (*expr_)[i_ + n];
    
      iterator& operator++() & { ++i_; }
      iterator operator++(int) & { auto tmp = *this; ++*this; return tmp; }
      // Similarly for operator--
    
      iterator operator+(difference_type n) const {
        return iterator{expr_, i_ + n};
      }
      // Similarly for operators -, +=, and -=
    
      friend iterator operator+(difference_type n, const iterator& i) {
        return i + n;
      }
    
    private:
        Expr* expr_;
        difference_type i_;
    };
    

    Now you simply have to arrange for "Expression templates" to have begin and end members that return iterator{*this, 0} and iterator{*this, size()}.