429 lines
12 KiB
C++
429 lines
12 KiB
C++
// bind_tests_advanced.cpp -- The Boost Lambda Library ------------------
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//
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// Copyright (C) 2000-2003 Jaakko Jarvi (jaakko.jarvi@cs.utu.fi)
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// Copyright (C) 2000-2003 Gary Powell (powellg@amazon.com)
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// Copyright (C) 2010 Steven Watanabe
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//
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// Distributed under the Boost Software License, Version 1.0. (See
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// accompanying file LICENSE_1_0.txt or copy at
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// http://www.boost.org/LICENSE_1_0.txt)
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//
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// For more information, see www.boost.org
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// -----------------------------------------------------------------------
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#include <boost/test/minimal.hpp> // see "Header Implementation Option"
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#include "boost/lambda/lambda.hpp"
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#include "boost/lambda/bind.hpp"
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#include "boost/any.hpp"
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#include "boost/type_traits/is_reference.hpp"
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#include "boost/mpl/assert.hpp"
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#include "boost/mpl/if.hpp"
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#include <iostream>
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#include <functional>
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#include <algorithm>
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using namespace boost::lambda;
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namespace bl = boost::lambda;
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int sum_0() { return 0; }
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int sum_1(int a) { return a; }
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int sum_2(int a, int b) { return a+b; }
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int product_2(int a, int b) { return a*b; }
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// unary function that returns a pointer to a binary function
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typedef int (*fptr_type)(int, int);
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fptr_type sum_or_product(bool x) {
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return x ? sum_2 : product_2;
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}
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// a nullary functor that returns a pointer to a unary function that
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// returns a pointer to a binary function.
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struct which_one {
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typedef fptr_type (*result_type)(bool x);
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template <class T> struct sig { typedef result_type type; };
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result_type operator()() const { return sum_or_product; }
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};
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void test_nested_binds()
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{
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int j = 2; int k = 3;
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// bind calls can be nested (the target function can be a lambda functor)
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// The interpretation is, that the innermost lambda functor returns something
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// that is bindable (another lambda functor, function pointer ...)
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bool condition;
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condition = true;
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BOOST_CHECK(bind(bind(&sum_or_product, _1), 1, 2)(condition)==3);
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BOOST_CHECK(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==5);
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condition = false;
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BOOST_CHECK(bind(bind(&sum_or_product, _1), 1, 2)(condition)==2);
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BOOST_CHECK(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==6);
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which_one wo;
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BOOST_CHECK(bind(bind(bind(wo), _1), _2, _3)(condition, j, k)==6);
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return;
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}
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// unlambda -------------------------------------------------
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// Sometimes it may be necessary to prevent the argument substitution of
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// taking place. For example, we may end up with a nested bind expression
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// inadvertently when using the target function is received as a parameter
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template<class F>
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int call_with_100(const F& f) {
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// bind(f, _1)(make_const(100));
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// This would result in;
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// bind(_1 + 1, _1)(make_const(100)) , which would be a compile time error
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return bl::bind(unlambda(f), _1)(make_const(100));
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// for other functors than lambda functors, unlambda has no effect
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// (except for making them const)
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}
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template<class F>
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int call_with_101(const F& f) {
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return bind(unlambda(f), _1)(make_const(101));
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}
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void test_unlambda() {
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int i = 1;
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BOOST_CHECK(unlambda(_1 + _2)(i, i) == 2);
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BOOST_CHECK(unlambda(++var(i))() == 2);
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BOOST_CHECK(call_with_100(_1 + 1) == 101);
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BOOST_CHECK(call_with_101(_1 + 1) == 102);
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#if defined(BOOST_NO_CXX11_HDR_FUNCTIONAL)
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BOOST_CHECK(call_with_100(bl::bind(std_functor(std::bind1st(std::plus<int>(), 1)), _1)) == 101);
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#else
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BOOST_CHECK(call_with_100(bl::bind(std_functor(std::bind(std::plus<int>(), 1, std::placeholders::_1)), _1)) == 101);
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#endif
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// std_functor insturcts LL that the functor defines a result_type typedef
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// rather than a sig template.
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bl::bind(std_functor(std::plus<int>()), _1, _2)(i, i);
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}
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// protect ------------------------------------------------------------
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// protect protects a lambda functor from argument substitution.
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// protect is useful e.g. with nested stl algorithm calls.
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namespace ll {
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struct for_each {
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// note, std::for_each returns it's last argument
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// We want the same behaviour from our ll::for_each.
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// However, the functor can be called with any arguments, and
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// the return type thus depends on the argument types.
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// 1. Provide a sig class member template:
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// The return type deduction system instantiate this class as:
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// sig<Args>::type, where Args is a boost::tuples::cons-list
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// The head type is the function object type itself
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// cv-qualified (so it is possilbe to provide different return types
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// for differently cv-qualified operator()'s.
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// The tail type is the list of the types of the actual arguments the
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// function was called with.
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// So sig should contain a typedef type, which defines a mapping from
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// the operator() arguments to its return type.
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// Note, that it is possible to provide different sigs for the same functor
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// if the functor has several operator()'s, even if they have different
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// number of arguments.
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// Note, that the argument types in Args are guaranteed to be non-reference
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// types, but they can have cv-qualifiers.
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template <class Args>
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struct sig {
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typedef typename boost::remove_const<
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typename boost::tuples::element<3, Args>::type
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>::type type;
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};
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template <class A, class B, class C>
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C
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operator()(const A& a, const B& b, const C& c) const
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{ return std::for_each(a, b, c);}
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};
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} // end of ll namespace
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void test_protect()
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{
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int i = 0;
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int b[3][5];
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int* a[3];
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for(int j=0; j<3; ++j) a[j] = b[j];
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std::for_each(a, a+3,
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bind(ll::for_each(), _1, _1 + 5, protect(_1 = ++var(i))));
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// This is how you could output the values (it is uncommented, no output
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// from a regression test file):
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// std::for_each(a, a+3,
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// bind(ll::for_each(), _1, _1 + 5,
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// std::cout << constant("\nLine ") << (&_1 - a) << " : "
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// << protect(_1)
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// )
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// );
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int sum = 0;
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std::for_each(a, a+3,
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bind(ll::for_each(), _1, _1 + 5,
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protect(sum += _1))
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);
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BOOST_CHECK(sum == (1+15)*15/2);
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sum = 0;
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std::for_each(a, a+3,
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bind(ll::for_each(), _1, _1 + 5,
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sum += 1 + protect(_1)) // add element count
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);
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BOOST_CHECK(sum == (1+15)*15/2 + 15);
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(1 + protect(_1))(sum);
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int k = 0;
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((k += constant(1)) += protect(constant(2)))();
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BOOST_CHECK(k==1);
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k = 0;
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((k += constant(1)) += protect(constant(2)))()();
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BOOST_CHECK(k==3);
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// note, the following doesn't work:
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// ((var(k) = constant(1)) = protect(constant(2)))();
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// (var(k) = constant(1))() returns int& and thus the
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// second assignment fails.
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// We should have something like:
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// bind(var, var(k) = constant(1)) = protect(constant(2)))();
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// But currently var is not bindable.
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// The same goes with ret. A bindable ret could be handy sometimes as well
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// (protect(std::cout << _1), std::cout << _1)(i)(j); does not work
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// because the comma operator tries to store the result of the evaluation
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// of std::cout << _1 as a copy (and you can't copy std::ostream).
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// something like this:
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// (protect(std::cout << _1), bind(ref, std::cout << _1))(i)(j);
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// the stuff below works, but we do not want extra output to
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// cout, must be changed to stringstreams but stringstreams do not
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// work due to a bug in the type deduction. Will be fixed...
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#if 0
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// But for now, ref is not bindable. There are other ways around this:
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int x = 1, y = 2;
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(protect(std::cout << _1), (std::cout << _1, 0))(x)(y);
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// added one dummy value to make the argument to comma an int
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// instead of ostream&
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// Note, the same problem is more apparent without protect
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// (std::cout << 1, std::cout << constant(2))(); // does not work
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(boost::ref(std::cout << 1), std::cout << constant(2))(); // this does
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#endif
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}
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void test_lambda_functors_as_arguments_to_lambda_functors() {
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// lambda functor is a function object, and can therefore be used
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// as an argument to another lambda functors function call object.
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// Note however, that the argument/type substitution is not entered again.
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// This means, that something like this will not work:
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(_1 + _2)(_1, make_const(7));
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(_1 + _2)(bind(&sum_0), make_const(7));
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// or it does work, but the effect is not to call
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// sum_0() + 7, but rather
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// bind(sum_0) + 7, which results in another lambda functor
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// (lambda functor + int) and can be called again
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BOOST_CHECK((_1 + _2)(bind(&sum_0), make_const(7))() == 7);
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int i = 3, j = 12;
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BOOST_CHECK((_1 - _2)(_2, _1)(i, j) == j - i);
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// also, note that lambda functor are no special case for bind if received
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// as a parameter. In oder to be bindable, the functor must
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// defint the sig template, or then
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// the return type must be defined within the bind call. Lambda functors
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// do define the sig template, so if the return type deduction system
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// covers the case, there is no need to specify the return type
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// explicitly.
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int a = 5, b = 6;
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// Let type deduction find out the return type
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BOOST_CHECK(bind(_1, _2, _3)(unlambda(_1 + _2), a, b) == 11);
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//specify it yourself:
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BOOST_CHECK(bind(_1, _2, _3)(ret<int>(_1 + _2), a, b) == 11);
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BOOST_CHECK(ret<int>(bind(_1, _2, _3))(_1 + _2, a, b) == 11);
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BOOST_CHECK(bind<int>(_1, _2, _3)(_1 + _2, a, b) == 11);
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bind(_1,1.0)(_1+_1);
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return;
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}
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void test_const_parameters() {
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// (_1 + _2)(1, 2); // this would fail,
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// Either make arguments const:
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BOOST_CHECK((_1 + _2)(make_const(1), make_const(2)) == 3);
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// Or use const_parameters:
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BOOST_CHECK(const_parameters(_1 + _2)(1, 2) == 3);
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}
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void test_rvalue_arguments()
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{
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// Not quite working yet.
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// Problems with visual 7.1
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// BOOST_CHECK((_1 + _2)(1, 2) == 3);
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}
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void test_break_const()
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{
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// break_const is currently unnecessary, as LL supports perfect forwarding
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// for up to there argument lambda functors, and LL does not support
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// lambda functors with more than 3 args.
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// I'll keep the test case around anyway, if more arguments will be supported
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// in the future.
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// break_const breaks constness! Be careful!
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// You need this only if you need to have side effects on some argument(s)
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// and some arguments are non-const rvalues and your lambda functors
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// take more than 3 arguments.
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int i = 1;
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// OLD COMMENT: (_1 += _2)(i, 2) // fails, 2 is a non-const rvalue
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// OLD COMMENT: const_parameters(_1 += _2)(i, 2) // fails, side-effect to i
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break_const(_1 += _2)(i, 2); // ok
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BOOST_CHECK(i == 3);
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}
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template<class T>
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struct func {
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template<class Args>
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struct sig {
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typedef typename boost::tuples::element<1, Args>::type arg1;
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// If the argument type is not the same as the expected type,
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// return void, which will cause an error. Note that we
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// can't just assert that the types are the same, because
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// both const and non-const versions can be instantiated
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// even though only one is ultimately used.
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typedef typename boost::mpl::if_<boost::is_same<arg1, T>,
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typename boost::remove_const<arg1>::type,
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void
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>::type type;
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};
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template<class U>
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U operator()(const U& arg) const {
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return arg;
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}
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};
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void test_sig()
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{
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int i = 1;
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BOOST_CHECK(bind(func<int>(), 1)() == 1);
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BOOST_CHECK(bind(func<const int>(), _1)(static_cast<const int&>(i)) == 1);
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BOOST_CHECK(bind(func<int>(), _1)(i) == 1);
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}
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class base {
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public:
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virtual int foo() = 0;
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};
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class derived : public base {
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public:
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virtual int foo() {
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return 1;
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}
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};
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void test_abstract()
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{
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derived d;
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base& b = d;
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BOOST_CHECK(bind(&base::foo, var(b))() == 1);
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BOOST_CHECK(bind(&base::foo, *_1)(&b) == 1);
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}
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int test_main(int, char *[]) {
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test_nested_binds();
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test_unlambda();
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test_protect();
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test_lambda_functors_as_arguments_to_lambda_functors();
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test_const_parameters();
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test_rvalue_arguments();
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test_break_const();
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test_sig();
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test_abstract();
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return 0;
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}
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