如何使 boost-proto 函数表达式可流化?
我正在查看 boost-proto 教程,并在使用惰性 pow 函数示例时遇到了这个问题。这是示例代码:
// Define a pow_fun function object
template<int Exp> // , typename Func>
struct pow_fun
{
typedef double result_type;
double operator()(double d) const
{
return pow(d, Exp);
}
};
// Define a lazy pow() function for the calculator DSEL.
// Can be used as: pow< 2 >(_1)
template<int Exp, typename Arg>
typename proto::result_of::make_expr<
proto::tag::function // Tag type
, pow_fun<Exp> // First child (by value)
, Arg const & // Second child (by reference)
>::type const
mypow(Arg const &arg)
{
return proto::make_expr<proto::tag::function>(
pow_fun<Exp>() // First child (by value)
, boost::ref(arg) // Second child (by reference)
);
}
现在,如果我尝试
proto::display_expr( mypow<2>(_1) );
编译器会抱怨它没有运算符<<为 函数表达式。我如何定义一个?
谢谢。
编译器错误为:
/usr/include/boost/proto/debug.hpp:146: error: no match for 'operator<<'在 'std::operator<< 中[with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits]((( std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [与 _CharT = char,_Traits = std::char_traits](((std::basic_ostream >&)((std::ostream*)((const boost::proto::function::display_expr*)this)->boost::proto::function::display_expr::sout_)), std ::setw(((常量boost::proto::function::display_expr*)this)->boost::proto::function::display_expr::深度_)))), (((const boost::proto::function::display_expr* )this)->boost::proto::function::display_expr::first_ ? ((const char*)"") : ((const char*)", "))))), boost::proto::tag::proto_tag_name((boost::proto::tag::terminal(), boost::proto::tag::terminal()))))), ((const char*)"(")) << boost::proto::value [with Expr = boost::proto::exprns_::expr >, 0l>](((const boost::proto::exprns_::expr >, 0l>&)((const boost::proto::exprns_::expr >, 0l>*)expr)))'
I'm going over the boost-proto tutorial, and ran into this problem with the lazy pow function example. This is the example code:
// Define a pow_fun function object
template<int Exp> // , typename Func>
struct pow_fun
{
typedef double result_type;
double operator()(double d) const
{
return pow(d, Exp);
}
};
// Define a lazy pow() function for the calculator DSEL.
// Can be used as: pow< 2 >(_1)
template<int Exp, typename Arg>
typename proto::result_of::make_expr<
proto::tag::function // Tag type
, pow_fun<Exp> // First child (by value)
, Arg const & // Second child (by reference)
>::type const
mypow(Arg const &arg)
{
return proto::make_expr<proto::tag::function>(
pow_fun<Exp>() // First child (by value)
, boost::ref(arg) // Second child (by reference)
);
}
Now, if I try to
proto::display_expr( mypow<2>(_1) );
the compiler complains that it doesn't have operator<< for the
function expression. How do I define one?
Thanks.
The compiler error is:
/usr/include/boost/proto/debug.hpp:146: error: no match for ‘operator<<’ in ‘std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _CharT = char, _Traits = std::char_traits](((std::basic_ostream >&)((std::ostream*)((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::sout_)), std::setw(((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::depth_)))), (((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::first_ ? ((const char*)"") : ((const char*)", "))))), boost::proto::tag::proto_tag_name((boost::proto::tag::terminal(), boost::proto::tag::terminal()))))), ((const char*)"(")) << boost::proto::value [with Expr = boost::proto::exprns_::expr >, 0l>](((const boost::proto::exprns_::expr >, 0l>&)((const boost::proto::exprns_::expr >, 0l>*)expr)))’
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这是哪个原型版本?最新的不需要 <<不再重载,如果需要,默认使用 typeid 来显示名称。您能发布实际的错误消息吗?
Which proto version is this ? The latest don't require the << overload anymore and default to typeid to display name if needed. Could you post the actual error message ?