当我使用以下测试代码尝试mpl::bind函数时，我无法在gcc中传递编译器，
有人可以帮助我找出问题所在，非常感谢。
#include <iostream>
#include <typeinfo>
#include <string>
#include <boost/mpl/apply.hpp>
#include <boost/mpl/char.hpp>
#include <boost/mpl/int.hpp>
#include <boost/mpl/arg.hpp>
#include <boost/mpl/plus.hpp>
#include <boost/mpl/placeholders.hpp>
#include <boost/static_assert.hpp>
#include <boost/type_traits/add_pointer.hpp>
#include <boost/type_traits/is_same.hpp>
#include <boost/mpl/quote.hpp>
using namespace std;
using namespace boost::mpl;
template< typename T1,typename T2 >
struct int_plus:boost::mpl::int_< (T1::value+T2::value) >
{
};
int main()
{
typedef boost::mpl::lambda< int_plus<_1, _2 > >::type test1; //-fine
// test2 define is causeing error
typedef boost::mpl::bind < int_plus<_1, _2 > > test2; //-error?
typedef boost::mpl::lambda< quote2<int_plus>, _2, _1 >::type test3; //-fine
typedef boost::mpl::bind< quote2<int_plus>, _2, _1 > test4; //-fine
typedef test1::apply<int_<42>, int_<23>>::type test5; //-fine
typedef test2::apply<int_<42>, int_<23>>::type test6; //-error
typedef test3::apply<int_<42>, int_<24>>::type test7; //-fine
typedef test4::apply<int_<42>, int_<24>>::type test8; //-fine
BOOST_MPL_ASSERT_RELATION( test5::value, ==, 65 ); //-fine
//BOOST_MPL_ASSERT_RELATION( test6::value, ==, 65 );
}
错误消息:
|| ===构建:在jtest2中调试(编译器:GNU GCC编译器)=== |
C:\ boost \ mpl \ aux_ \ preprocessed \ gcc \ apply_wrap.hpp ||实例化为'struct boost::mpl::apply_wrap0，mpl _::arg >，mpl _::bool_>':|
C:\ boost \ mpl \ aux_ \ preprocessed \ gcc \ bind.hpp | 86 |从'struct boost::mpl::bind0，mpl _::arg >>::apply，mpl _::int_ >'|
C:\ ls \ jtest2 \ main.cpp | 30 |从此处要求|
C:\ boost \ mpl \ aux_ \ preprocessed \ gcc \ apply_wrap.hpp | 20 |错误:在'struct int_plus，mpl _::arg >'|中没有名为'apply'的类模板
C:\ boost \ mpl \ aux_ \ preprocessed \ gcc \ bind.hpp ||在'struct boost::mpl::bind0，mpl _::arg >>::apply，mpl _::int_ >':|
C:\ ls \ jtest2 \ main.cpp | 30 |从此处要求|
C:\ boost \ mpl \ aux_ \ preprocessed \ gcc \ bind.hpp | 86 |错误:在'struct boost::mpl::apply_wrap0，mpl _::arg >，mpl_::中没有名为'type'的类型bool_>'|
|| ===构建失败:2个错误，5个警告(0分钟，0秒)=== |
最佳答案
在检查bind的定义和语义之后，它需要一个元函数类作为第一个参数，这意味着该元函数无法工作；
在此示例中，我们有几种方法可以将元函数转换为matafunction类
元函数int_plus可以被隐藏
1)quote2(int_plus)
2)int_plus_f
3)int_plus_f2
#include <iostream>
#include <typeinfo>
#include <string>
#include <boost/mpl/apply.hpp>
#include <boost/mpl/char.hpp>
#include <boost/mpl/int.hpp>
#include <boost/mpl/arg.hpp>
#include <boost/mpl/plus.hpp>
#include <boost/mpl/placeholders.hpp>
#include <boost/static_assert.hpp>
#include <boost/type_traits/add_pointer.hpp>
#include <boost/type_traits/is_same.hpp>
#include <boost/mpl/quote.hpp>
using namespace std;
using namespace boost::mpl;
template< typename T1,typename T2 >
struct int_plus:boost::mpl::int_< (T1::value+T2::value) >
{
};
struct int_plus_f // method 1 to get metafunction class, not perfect for lambda
{
template< typename T1,typename T2 >
struct apply:boost::mpl::int_< (T1::value+T2::value) >
{
};
};
struct int_plus_f2 // method 2 to get metafunction class, perfect for lambda
{
template< typename A1, typename A2 > struct apply
: int_plus<A1,A2>
{
};
};
int main()
{
//bind define:
// typedef bind<f,a1,...an> g;
//bind parameters:
// F Metafunction Class An metafunction class to perform binding on.
// A1,... An Any type Arguments to bind.
//lambda define:
// typedef lambda<x>::type f;
// typedef lambda<x,Tag>::type f;
//lambda parameters
// X Any type An expression to transform.
// Tag Any type A tag determining transform semantics
//lambda Semantics equivalent to
// typedef protect< bind< quoten<X> , lambda<a1>::type,... lambda<an>::type > > f;
//quote define:
// typedef quoten<f> g;
// typedef quoten<f,tag> g;
//quote2 Semantics Equivalent to
// struct g{
// template< typename A1,typename A2 >
// struct apply : f<A1,A2>{};
// };
typedef boost::mpl::lambda< int_plus<_1, _2 > >::type test1; //-fine
typedef boost::mpl::bind < int_plus_f,_1, _2 > test2; //-fine
typedef boost::mpl::bind < int_plus_f2,_1, _2 > test3; //-fine
typedef boost::mpl::lambda< int_plus_f2,_1, _2 >::type test4; //-fine
typedef boost::mpl::lambda< quote2<int_plus>, _2, _1 >::type test5; //-fine
typedef boost::mpl::bind< quote2<int_plus>, _2, _1 > test6; //-fine
typedef test1::apply<int_<42>, int_<22>>::type result1; //-fine
typedef test2::apply<int_<42>, int_<23>>::type result2; //-fine
typedef test3::apply<int_<42>, int_<24>>::type result3; //-fine
typedef test4::apply<int_<42>, int_<25>>::type result4; //-fine
typedef test5::apply<int_<42>, int_<26>>::type result5; //-fine
typedef test6::apply<int_<42>, int_<27>>::type result6; //-fine
BOOST_MPL_ASSERT_RELATION( result1::value, ==, 64 ); //-fine
BOOST_MPL_ASSERT_RELATION( result2::value, ==, 65 ); //-fine
BOOST_MPL_ASSERT_RELATION( result3::value, ==, 66 ); //-fine
BOOST_MPL_ASSERT_RELATION( result4::value, ==, 67 ); //-fine
BOOST_MPL_ASSERT_RELATION( result5::value, ==, 68 ); //-fine
BOOST_MPL_ASSERT_RELATION( result6::value, ==, 69 ); //-fine
//apply : Invokes a Metafunction Class or a Lambda Expression F with arguments A1,... An.
// typedef apply<f,a1,...an>::type t;
//apply parameters
// F Lambda Expression: An expression(e.g.: a metafunction) to invoke,
// metafunction class is fine also
// A1,... An Any type Invocation arguments.
// apply Semantics Equivalent to
// typedef apply_wrapn< lambda<f>::type,a1,... an>::type t;.
typedef apply< int_plus<_1,_2>, int_<2>, int_<3> >::type r1;
typedef apply< quote2<int_plus>, int_<2>, int_<3> >::type r2;
typedef apply< int_plus_f, int_<2>, int_<3> >::type r3;
typedef apply< int_plus_f2, int_<2>, int_<3> >::type r4;
BOOST_MPL_ASSERT_RELATION( r1::value, ==, 5 );
BOOST_MPL_ASSERT_RELATION( r2::value, ==, 5 );
BOOST_MPL_ASSERT_RELATION( r3::value, ==, 5 );
BOOST_MPL_ASSERT_RELATION( r4::value, ==, 5 );
}
关于c++ - boost::mpl::bind的用法，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/22324233/