// bind_tests_advanced.cpp -------------------------------- #define BOOST_INCLUDE_MAIN // for testing, include rather than link #include // see "Header Implementation Option" #include "boost/lambda/bind.hpp" #include "boost/lambda/lambda.hpp" #include "boost/any.hpp" #include #include #include using namespace boost::lambda; int sum_0() { return 0; } int sum_1(int a) { return a; } int sum_2(int a, int b) { return a+b; } int product_2(int a, int b) { return a*b; } // unary function that returns a pointer to a binary function typedef int (*fptr_type)(int, int); fptr_type sum_or_product(bool x) { return x ? sum_2 : product_2; } // a nullary functor that returns a pointer to a unary function that // returns a pointer to a binary function. struct which_one { typedef fptr_type (*result_type)(bool x); result_type operator()() const { return sum_or_product; } }; void test_nested_binds() { int j = 2; int k = 3; // bind calls can be nested (the target function can be a lambda functor) // The interpretation is, that the innermost lambda functor returns something // that is bindable (another lambda functor, function pointer ...) bool condition; condition = true; BOOST_TEST(bind(bind(&sum_or_product, _1), 1, 2)(condition)==3); BOOST_TEST(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==5); condition = false; BOOST_TEST(bind(bind(&sum_or_product, _1), 1, 2)(condition)==2); BOOST_TEST(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==6); which_one wo; BOOST_TEST(bind(bind(bind(wo), _1), _2, _3)(condition, j, k)==6); return; } // unlambda ------------------------------------------------- // Sometimes it may be necessary to prevent the argument substitution of // taking place. For example, we may end up with a nested bind expression // inadvertently when using the target function is received as a parameter template int call_with_100(const F& f) { // bind(f, _1)(make_const(100)); // This would result in; // bind(_1 + 1, _1)(make_const(100)) , which would be a compile time error return bind(unlambda(f), _1)(make_const(100)); // for other functors than lambda functors, unlambda has no effect // (except for making them const) } template int call_with_101(const F& f) { return bind(unlambda(ret(f)), _1)(make_const(101)); // the ret must be inside of unlambda, since unlambda requires its argument // to define result_type. // if F is not a lambda functor ret(f) fails at compile time! } void test_unlambda() { BOOST_TEST(call_with_100(ret(_1 + 1)) == 101); // note, that the functor must define the result_type typedef, as the bind // int the called function does not do that. BOOST_TEST(call_with_101(_1 + 1) == 102); // This one leaves the return type to be specified by the bind in the // called function, and that makes things kind of hard in the called // function BOOST_TEST(call_with_100(std::bind1st(std::plus(), 1)) == 101); // BOOST_TEST(call_with_101(std::bind1st(std::plus(), 1)) == 102); // this would fail, as it would lead to ret being called with other than // a lambda functor } // protect ------------------------------------------------------------ // protect protects a lambda functor from argument substitution. // protect is useful e.g. with nested stl algorithm calls. namespace ll { struct for_each : public has_sig { // note, std::for_each returns it's last argument // We want the same behaviour from our ll::for_each. // However, the functor can be called with any arguments, and // the return type thus depends on the argument types. // The basic mechanism (provide a result_type typedef) does not // work. // There is an alternative for this kind of situations, which LL // borrows from FC++ (by Yannis Smaragdakis and Brian McNamara). // If you want to use this mechanism, your function object class needs to // 1. inhertit publicly from has_sig // 2. Provide a sig class member template: // The return type deduction system instantiate this class as: // sig::type, where Args is a boost::tuples::cons-list // The head type is the function object type itself // cv-qualified (so it is possilbe to provide different return types // for differently cv-qualified operator()'s. // The tail type is the list of the types of the actual arguments the // function was called with. // So sig should contain a typedef type, which defines a mapping from // the operator() arguments to its return type. // Note, that it is possible to provide different sigs for the same functor // if the functor has several operator()'s, even if they have different // number of arguments. // Note, that the argument types in Args can be arbitrary types, particularly // they can be reference types and can have qualifiers or both. // So some care will be needed in this respect. template struct sig { typedef typename boost::remove_const< typename boost::remove_reference< typename boost::tuples::element<3, Args>::type >::type >::type type; }; template C operator()(const A& a, const B& b, const C& c) const { return std::for_each(a, b, c);} }; } // end of ll namespace void test_protect() { int i = 0; int b[3][5]; int* a[3]; for(int j=0; j<3; ++j) a[j] = b[j]; std::for_each(a, a+3, bind(ll::for_each(), _1, _1 + 5, protect(_1 = ++var(i)))); // This is how you could output the values (it is uncommented, no output // from a regression test file): // std::for_each(a, a+3, // bind(ll::for_each(), _1, _1 + 5, // std::cout << constant("\nLine ") << (&_1 - a) << " : " // << protect(_1) // ) // ); int sum = 0; std::for_each(a, a+3, bind(ll::for_each(), _1, _1 + 5, protect(sum += _1)) ); BOOST_TEST(sum == (1+15)*15/2); sum = 0; std::for_each(a, a+3, bind(ll::for_each(), _1, _1 + 5, sum += 1 + protect(_1)) // add element count ); BOOST_TEST(sum == (1+15)*15/2 + 15); int k = 0; ((k += constant(1)) += protect(constant(2)))(); BOOST_TEST(k==1); k = 0; ((k += constant(1)) += protect(constant(2)))()(); BOOST_TEST(k==3); // note, the following doesn't work: // ((var(k) = constant(1)) = protect(constant(2)))(); // (var(k) = constant(1))() returns int& and thus the // second assignment fails. // We should have something like: // bind(var, var(k) = constant(1)) = protect(constant(2)))(); // But currently var is not bindable. // The same goes with ret. A bindable ret could be handy sometimes as well // (protect(std::cout << _1), std::cout << _1)(i)(j); does not work // because the comma operator tries to store the result of the evaluation // of std::cout << _1 as a copy (and you can't copy std::ostream). // something like this: // (protect(std::cout << _1), bind(ref, std::cout << _1))(i)(j); // But for now, ref is not bindable. There are other ways around this: // int x = 1, y = 2; // (protect(std::cout << _1), (std::cout << _1, 0))(x)(y); // added one dummy value to make the argument to comma an int // instead of ostream& // Note, the same problem is more apparent without protect // (std::cout << 1, std::cout << constant(2))(); // does not work // (boost::ref(std::cout << 1), std::cout << constant(2))(); // this does } void test_lambda_functors_as_arguments_to_lambda_functors() { // lambda functor is a function object, and can therefore be used // as an argument to another lambda functors function call object. // Note however, that the argument/type substitution is not entered again. // This means, that something like this will not work: (_1 + _2)(bind(&sum_0), make_const(7)); // or it does work, but the effect is not to call // sum_0() + 7, but rather // bind(sum_0) + 7, which results in another lambda functor // (lambda functor + int) and can be called again BOOST_TEST((_1 + _2)(bind(&sum_0), make_const(7))() == 7); // also, note that lambda functor are no special case for bind if received // as a parameter. In oder to be bindable, the functor must // either define the result_type typedef, have the sig template, or then // the return type must be defined within the bind call. Lambda functors // do define the sig template, so if the return type deduction system // covers the case, there is no need to specify the return type // explicitly. int a = 5, b = 6; // Let type deduction take find out the return type BOOST_TEST(bind(_1, _2, _3)(_1 + _2, a, b) == 11); //specify it yourself: BOOST_TEST(bind(_1, _2, _3)(ret(_1 + _2), a, b) == 11); BOOST_TEST(ret(bind(_1, _2, _3))(_1 + _2, a, b) == 11); BOOST_TEST(bind(_1, _2, _3)(_1 + _2, a, b) == 11); return; } void test_currying() { int a = 1, b = 2, c = 3; // lambda functors support currying: // binary functor can be called with just one argument, the result is // a unary lambda functor. // 3-ary functor can be called with one or two arguments (and with 3 // of course) BOOST_TEST((_1 + _2)(a)(b) == 3); BOOST_TEST((_1 + _2 + _3)(a, b)(c) == 6); BOOST_TEST((_1 + _2 + _3)(a)(b, c) == 6); BOOST_TEST((_1 + _2 + _3)(a)(b)(c) == 6); // Also, lambda functors passed as arguments end up being curryable BOOST_TEST(bind(_1, _2, _3)(_1 + _2 + _3, a, b)(c) == 6); BOOST_TEST(bind(_1, _2)(_1 + _2 + _3, a)(b, c) == 6); BOOST_TEST(bind(_1, _2)(_1 + _2 + _3, a)(b)(c) == 6); bind(_1, _2)(_1 += (_2 + _3), a)(b)(c); BOOST_TEST(a == 6); bind(_1, _2)(a += (_1 + _2 + _3), c)(c)(c); BOOST_TEST(a == 6+3*c); a = 1, b = 2, c = 3; // and protecting should work as well BOOST_TEST(bind(_1, _2)(_1 + _2 + _3 + protect(_1), a)(b)(c)(a) == 7); return; } void test_const_parameters() { // (_1 + _2)(1, 2); // this would fail, // Either make arguments const: BOOST_TEST((_1 + _2)(make_const(1), make_const(2)) == 3); // Or use const_parameters: BOOST_TEST(const_parameters(_1 + _2)(1, 2) == 3); } void test_break_const() { // break_const breaks constness! Be careful! // You need this only if you need to have side effects on some argument(s) // and some arguments are non-const rvalues: // E.g. int i = 1; // (_1 += _2)(i, 2) // fails, 2 is a non-const rvalue // const_parameters(_1 += _2)(i, 2) // fails, side-effect to i break_const(_1 += _2)(i, 2); // ok BOOST_TEST(i == 3); } int test_main(int, char *[]) { test_nested_binds(); test_unlambda(); test_protect(); test_lambda_functors_as_arguments_to_lambda_functors(); test_currying(); test_const_parameters(); test_break_const(); return 0; }