为什么不自动向下转换应用于模板功能? [英] Why isn't automatic downcasting applied to template functions?

问题描述

有人问此问题有关字符串附加的问题. string s; s = s + 2;未编译.人们给出了答案，指出operator+被定义为模板函数，而operator+=未定义为模板函数，因此不应用自动向下转换(int(2)至char(2)).

Someone asked this question about string appending. It's string s; s = s + 2; not compiling. People gave answers stating that operator+ is defined as a template function while operator+= is not, so auto downcasting (int(2) to char(2)) is not applied.

原型是

template<typename _CharT, typename _Traits, typename _Alloc>
class basic_string{
    basic_string&
      operator+=(_CharT __c);
};

template<typename _CharT, typename _Traits, typename _Alloc>
  inline basic_string<_CharT, _Traits, _Alloc>
  operator+(const basic_string<_CharT, _Traits, _Alloc>& __lhs, _CharT __rhs);

为什么编译器不能仅使用此原型并将int(2)转换为char(2)?

Why can't the compiler just use this prototype and cast int(2) to char(2)?

basic_string<char, _T, _A> operator+(const basic_string<char, _T, _A>, char);

编译器(G ++ 6.3.0)抱怨

The compiler (G++ 6.3.0) complains that

[Note] deduced conflicting types for parameter '_CharT' ('char' and 'int')

推荐答案

关键区别在于，对于operator +=变体，std::basic_string的char类型模板参数，以及其RHS的参数类型为已经固定为char，而operator+模板必须根据其参数推论得出.

The key difference is that for the operator += variant, the char type template argument for the std::basic_string, and thus the argument type for its RHS, is already fixed to char, while the operator+ template has to deduce that from its arguments.

因此，对于+=情况，编译器知道您想要" int-> char转换，没有什么可推论的.

Thus, for the += case, the compiler knows you "want" the int->char conversion, there is nothing to deduce there.

另一方面，对于operator+情况，编译器正在查看模板

For the operator+ case on the other hand, the compiler is looking at the template

template<class CharT, class Traits, class Alloc>
    basic_string<CharT,Traits,Alloc>
        operator+( const basic_string<CharT,Traits,Alloc>& lhs,
                   CharT rhs );

，并且，当尝试确定CharT应该是什么时，它从第一个操作数获取CharT = char(因为std::string为std::basic_string<char>)，而从第二个操作数获取CharT = int.根据标准，这种冲突被定义为编译错误.

and, when trying to determine what CharT is supposed to be, it gets CharT = char from the first operand (as std::string is std::basic_string<char>) and CharT = int from the second operand. Such a conflict is defined to be a compilation error by the standard.

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