lambda/test/bind_tests_advanced.cpp

429 lines
12 KiB
C++

// bind_tests_advanced.cpp -- The Boost Lambda Library ------------------
//
// Copyright (C) 2000-2003 Jaakko Jarvi (jaakko.jarvi@cs.utu.fi)
// Copyright (C) 2000-2003 Gary Powell (powellg@amazon.com)
// Copyright (C) 2010 Steven Watanabe
//
// Distributed under the Boost Software License, Version 1.0. (See
// accompanying file LICENSE_1_0.txt or copy at
// http://www.boost.org/LICENSE_1_0.txt)
//
// For more information, see www.boost.org
// -----------------------------------------------------------------------
#include <boost/test/minimal.hpp> // see "Header Implementation Option"
#include "boost/lambda/lambda.hpp"
#include "boost/lambda/bind.hpp"
#include "boost/any.hpp"
#include "boost/type_traits/is_reference.hpp"
#include "boost/mpl/assert.hpp"
#include "boost/mpl/if.hpp"
#include <iostream>
#include <functional>
#include <algorithm>
using namespace boost::lambda;
namespace bl = 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);
template <class T> struct sig { typedef result_type type; };
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_CHECK(bind(bind(&sum_or_product, _1), 1, 2)(condition)==3);
BOOST_CHECK(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==5);
condition = false;
BOOST_CHECK(bind(bind(&sum_or_product, _1), 1, 2)(condition)==2);
BOOST_CHECK(bind(bind(&sum_or_product, _1), _2, _3)(condition, j, k)==6);
which_one wo;
BOOST_CHECK(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<class F>
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 bl::bind(unlambda(f), _1)(make_const(100));
// for other functors than lambda functors, unlambda has no effect
// (except for making them const)
}
template<class F>
int call_with_101(const F& f) {
return bind(unlambda(f), _1)(make_const(101));
}
void test_unlambda() {
int i = 1;
BOOST_CHECK(unlambda(_1 + _2)(i, i) == 2);
BOOST_CHECK(unlambda(++var(i))() == 2);
BOOST_CHECK(call_with_100(_1 + 1) == 101);
BOOST_CHECK(call_with_101(_1 + 1) == 102);
#if defined(BOOST_NO_CXX11_HDR_FUNCTIONAL)
BOOST_CHECK(call_with_100(bl::bind(std_functor(std::bind1st(std::plus<int>(), 1)), _1)) == 101);
#else
BOOST_CHECK(call_with_100(bl::bind(std_functor(std::bind(std::plus<int>(), 1, std::placeholders::_1)), _1)) == 101);
#endif
// std_functor insturcts LL that the functor defines a result_type typedef
// rather than a sig template.
bl::bind(std_functor(std::plus<int>()), _1, _2)(i, i);
}
// protect ------------------------------------------------------------
// protect protects a lambda functor from argument substitution.
// protect is useful e.g. with nested stl algorithm calls.
namespace ll {
struct for_each {
// 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.
// 1. Provide a sig class member template:
// The return type deduction system instantiate this class as:
// sig<Args>::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 are guaranteed to be non-reference
// types, but they can have cv-qualifiers.
template <class Args>
struct sig {
typedef typename boost::remove_const<
typename boost::tuples::element<3, Args>::type
>::type type;
};
template <class A, class B, class C>
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_CHECK(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_CHECK(sum == (1+15)*15/2 + 15);
(1 + protect(_1))(sum);
int k = 0;
((k += constant(1)) += protect(constant(2)))();
BOOST_CHECK(k==1);
k = 0;
((k += constant(1)) += protect(constant(2)))()();
BOOST_CHECK(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);
// the stuff below works, but we do not want extra output to
// cout, must be changed to stringstreams but stringstreams do not
// work due to a bug in the type deduction. Will be fixed...
#if 0
// 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
#endif
}
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)(_1, make_const(7));
(_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_CHECK((_1 + _2)(bind(&sum_0), make_const(7))() == 7);
int i = 3, j = 12;
BOOST_CHECK((_1 - _2)(_2, _1)(i, j) == j - i);
// also, note that lambda functor are no special case for bind if received
// as a parameter. In oder to be bindable, the functor must
// defint 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 find out the return type
BOOST_CHECK(bind(_1, _2, _3)(unlambda(_1 + _2), a, b) == 11);
//specify it yourself:
BOOST_CHECK(bind(_1, _2, _3)(ret<int>(_1 + _2), a, b) == 11);
BOOST_CHECK(ret<int>(bind(_1, _2, _3))(_1 + _2, a, b) == 11);
BOOST_CHECK(bind<int>(_1, _2, _3)(_1 + _2, a, b) == 11);
bind(_1,1.0)(_1+_1);
return;
}
void test_const_parameters() {
// (_1 + _2)(1, 2); // this would fail,
// Either make arguments const:
BOOST_CHECK((_1 + _2)(make_const(1), make_const(2)) == 3);
// Or use const_parameters:
BOOST_CHECK(const_parameters(_1 + _2)(1, 2) == 3);
}
void test_rvalue_arguments()
{
// Not quite working yet.
// Problems with visual 7.1
// BOOST_CHECK((_1 + _2)(1, 2) == 3);
}
void test_break_const()
{
// break_const is currently unnecessary, as LL supports perfect forwarding
// for up to there argument lambda functors, and LL does not support
// lambda functors with more than 3 args.
// I'll keep the test case around anyway, if more arguments will be supported
// in the future.
// 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 and your lambda functors
// take more than 3 arguments.
int i = 1;
// OLD COMMENT: (_1 += _2)(i, 2) // fails, 2 is a non-const rvalue
// OLD COMMENT: const_parameters(_1 += _2)(i, 2) // fails, side-effect to i
break_const(_1 += _2)(i, 2); // ok
BOOST_CHECK(i == 3);
}
template<class T>
struct func {
template<class Args>
struct sig {
typedef typename boost::tuples::element<1, Args>::type arg1;
// If the argument type is not the same as the expected type,
// return void, which will cause an error. Note that we
// can't just assert that the types are the same, because
// both const and non-const versions can be instantiated
// even though only one is ultimately used.
typedef typename boost::mpl::if_<boost::is_same<arg1, T>,
typename boost::remove_const<arg1>::type,
void
>::type type;
};
template<class U>
U operator()(const U& arg) const {
return arg;
}
};
void test_sig()
{
int i = 1;
BOOST_CHECK(bind(func<int>(), 1)() == 1);
BOOST_CHECK(bind(func<const int>(), _1)(static_cast<const int&>(i)) == 1);
BOOST_CHECK(bind(func<int>(), _1)(i) == 1);
}
class base {
public:
virtual int foo() = 0;
};
class derived : public base {
public:
virtual int foo() {
return 1;
}
};
void test_abstract()
{
derived d;
base& b = d;
BOOST_CHECK(bind(&base::foo, var(b))() == 1);
BOOST_CHECK(bind(&base::foo, *_1)(&b) == 1);
}
int test_main(int, char *[]) {
test_nested_binds();
test_unlambda();
test_protect();
test_lambda_functors_as_arguments_to_lambda_functors();
test_const_parameters();
test_rvalue_arguments();
test_break_const();
test_sig();
test_abstract();
return 0;
}