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- // -*- mode: c++; coding: utf-8 -*-
- // ra-ra/test - Using nested arrays as if they were arrays if higher rank.
- // (c) Daniel Llorens - 2014
- // This library is free software; you can redistribute it and/or modify it under
- // the terms of the GNU Lesser General Public License as published by the Free
- // Software Foundation; either version 3 of the License, or (at your option) any
- // later version.
- // TODO Make more things work and work efficiently.
- using std::cout, std::endl, std::flush, std::cerr, ra::TestRecorder;
- template <class T, int N> using Array = ra::Big<T, N>;
- template <class T> using Vec = Array<T, 1>;
- int main()
- {
- TestRecorder tr;
- tr.section("operators on nested arrays");
- {
- // default init is required to make vector-of-vector, but I still want Vec {} to mean 'an empty vector' and not a default-init vector.
- {
- auto c = Vec<int> {};
- tr.test_eq(0, c.len(0));
- tr.test_eq(0, c.size());
- }
- {
- auto c = Vec<Vec<int>> { {} , {1}, {1, 2} };
- std::ostringstream os;
- os << c;
- std::istringstream is(os.str());
- Vec<Vec<int>> d;
- is >> d;
- tr.test_eq(3, d.size());
- tr.test_eq(d[0], Vec<int>{});
- tr.test_eq(d[1], Vec<int>{1});
- tr.test_eq(d[2], Vec<int>{1, 2});
- // TODO Actually nested 'as if higher rank' should allow just (every(c==d)). This is explicit nesting.
- tr.test(every(ra::expr([](auto & c, auto & d) { return every(c==d); }, c.iter(), d.iter())));
- }
- }
- tr.section("selector experiments");
- {
- // These is an investigation of how to make a(ra::all, i) or a(i, ra::all) work.
- // The problem with a(ra::all, i) here is that we probably want to leave the iteration on ra::all for last. Otherwise the indexing is redone for each rank-1 cell.
- Vec<int> i = {0, 3, 1, 2};
- Array<double, 2> a({4, 4}, ra::_0-ra::_1);
- Array<double, 2> b = from([](auto && a, auto && i) -> decltype(auto) { return a(i); }, a.iter<1>(), start(i));
- tr.test_eq(a(0, i), b(0));
- tr.test_eq(a(1, i), b(1));
- tr.test_eq(a(2, i), b(2));
- tr.test_eq(a(3, i), b(3));
- // The problem with a(i) = a(i, ra::all) is that a(i) returns a nested expression, so it isn't equivalent to a(i, [0 1 ...]), and if we want to write it as a rank 2 expression, we can't use from() as above because the iterator we want is a(i).iter(), it depends on i.
- // So ...
- }
- tr.section("copying btw arrays nested in the same way");
- {
- Vec<ra::Small<int, 2>> a {{1, 2}, {3, 4}, {5, 6}};
- ra::Small<ra::Small<int, 2>, 3> b = a;
- tr.test_eq(ra::Small<int, 2> {1, 2}, b(0));
- tr.test_eq(ra::Small<int, 2> {3, 4}, b(1));
- tr.test_eq(ra::Small<int, 2> {5, 6}, b(2));
- b(0) = ra::Small<int, 2> {7, 9};
- b(1) = ra::Small<int, 2> {3, 4};
- b(2) = ra::Small<int, 2> {1, 6};
- a = b;
- tr.test_eq(ra::Small<int, 2> {7, 9}, a(0));
- tr.test_eq(ra::Small<int, 2> {3, 4}, a(1));
- tr.test_eq(ra::Small<int, 2> {1, 6}, a(2));
- }
- tr.section("TODO copying btw arrays nested in different ways");
- {
- Vec<ra::Small<int, 2>> a {{1, 2}, {3, 4}, {5, 6}};
- Array<int, 2> b({3, 2}, {1, 2, 3, 4, 5, 6});
- // there's one level of matching so a(0) matches to b(0, 0) and b(0, 1), a(1) to b(1, 0) and b(1, 1), etc. So this results in a(0) = 1 overwritten with a(0) = 2, etc. finally a = [[2, 2], [4, 4], [6, 6]]. Probably not what we want.
- a = b;
- // b = a; // dnc bc [x, y] ← z is ok but z ← [x, y] is not.
- cout << a << endl;
- }
- return tr.summary();
- }
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