隐式强制转换不适用于BOOST_STRONG_TYPEDEF和BOOST_SPIRIT_DEBUG_NODE [英] Implicit cast doesn't work for BOOST_STRONG_TYPEDEF and BOOST_SPIRIT_DEBUG_NODE

问题描述

我已经定义了boost :: spirit :: qi规则:

I have defined a boost::spirit::qi rule:

boost::spirit::qi::rule<Iterator, Identifier()> id;

其中标识符由以下位置定义:

where Identifier is defined by:

BOOST_STRONG_TYPEDEF(std::string, Identifier)

但是当我使用

BOOST_SPIRIT_DEBUG_NODE(id);

它无法编译并出现以下错误:

It fails to compile with following error:

boost_1_51_0/boost/spirit/home/support/attributes.hpp:1203: error: no match for 'operator<<' in 'out << val'

并列出了ostream的重载运算符.

and it lists the overloaded operators of ostream.

知道BOOST_STRONG_TYPEDEF已将演员转换为原始类型时，不应 使用operator<<时，编译器从标识符隐式转换为std :: string?还是有限制阻止编译器在尝试与其他运算符(即operator<<)匹配时应用某种类型的强制转换运算符?

Knowing that BOOST_STRONG_TYPEDEF defines a cast operator to the original type, shouldn't compiler implicitly cast from Identifier to std::string when using operator<<? or is there a restriction that prevents compiler to apply a cast operator of a type when it is trying to match the other operator (namely operator<< )?

当我定义以下运算符时，它将进行编译:

When I define the following operator it compiles:

inline std::ostream& operator<<(std::ostream& os, const Identifier& id)
{
    return os << static_cast<std::string const&>(id);
}

我正在使用gcc4.4.2

I am using gcc4.4.2

推荐答案

这与boost，strong_typedef或spirit无关.

This has nothing to do with boost, strong_typedef or spirit.

与模板参数的类型推导有很多关系.简而言之，当推导参数类型时，从不进行隐式转换[1]

It has a lot to do with type deduction for template arguments. In short, when argument types are deduced, implicit conversions never take place [1]

cf.

#include <iostream>
#include <string>
#include <boost/strong_typedef.hpp>

BOOST_STRONG_TYPEDEF(double, X)
int main() { std::cout << X(); }

没问题！将double替换为std::string，它将不再起作用.有什么不同?

No problem! Replace double by std::string, and it doesn't work anymore. What's different?

流操作符的声明不同.

对比度

ostream& ostream::operator<<(double);

收件人

template<typename _CharT, typename _Traits, typename _Alloc>
   inline basic_ostream<_CharT, _Traits>&
   operator<<(basic_ostream<_CharT, _Traits>&, basic_string<_CharT, _Traits, _Alloc> const&)

运算符重载是功能模板的事实不允许任何隐式转换.

The fact that the operator overload is a function template disallows any implicit conversions.

[1]我猜想initializer_list在这里可能看起来有点例外，它可以通过加宽/缩小来实现.虽然主题不同

[1] I guess initializer_list may look like a bit of an exception here, what with widening/narrowing that it can do. Different subject, though

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