模板推导与隐式用户定义的转换运算符

时间:2018-04-04 08:32:29

标签: c++ templates implicit-conversion c++-concepts c++20

我尝试实现一个涉及模板的用户定义类型转换的小例子。

#include <cassert>
#include <cstdint>
#include <iostream>
#include <stdexcept>
#include <type_traits>

template <typename T>
concept bool UIntegral = requires() {
    std::is_integral_v<T> && !std::is_signed_v<T>;
};

class Number
{
public:
    Number(uint32_t number): _number(number)
    {
        if (number == 1) {
            number = 0;
        }

        for (; number > 1; number /= 10);
        if (number == 0) {
            throw std::logic_error("scale must be a factor of 10");
        }
    }

    template <UIntegral T>
    operator T() const
    {
        return static_cast<T>(this->_number);
    }

private:
    uint32_t _number;
};

void changeScale(uint32_t& magnitude, Number scale)
{
    //magnitude *= scale.operator uint32_t();
    magnitude *= scale;
}

int main()
{
    uint32_t something = 5;
    changeScale(something, 100);
    std::cout << something << std::endl;

    return 0;
}

我收到以下编译错误(来自GCC 7.3.0):

  

main.cpp:在函数'void changeScale(uint32_t&amp;,Number)'中:

     

main.cpp:40:15:错误:'operator * ='不匹配(操作数类型为'uint32_t {aka unsigned int}'和'Number')

     

幅度* =比例;

注意该行已注释掉 - 这个有效:

//magnitude *= scale.operator uint32_t();

为什么不能自动推导模板化转换运算符?在此先感谢您的帮助。

[编辑]

我遵循删除概念的建议来使用Clang并查看其错误消息。我得到了以下内容(这是截断但足够的):

main.cpp:34:15: error: use of overloaded operator '*=' is ambiguous (with operand types 'uint32_t'
  (aka 'unsigned int') and 'Number')
magnitude *= scale;
~~~~~~~~~ ^  ~~~~~
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, float)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, double)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, long double)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, __float128)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, int)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, long long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, __int128)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned int)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned long long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned __int128)

因此,在开启概念之后,我假设强制转换数字的唯一方法是将其用于无符号整数类型 - 那么为什么编译器不能推断转换呢?

1 个答案:

答案 0 :(得分:1)

requires概念表达式的工作方式与SFINAE类似,它仅检查表达式有效,但不对其进行评估。

要让概念实际上 T限制为无符号整数类型,请使用bool表达式:

template<typename T>
concept bool UIntegral = std::is_integral_v<T> && !std::is_signed_v<T>;

这会解决你的问题吗?不幸的是,请继续阅读...

  

为什么不能自动推导模板化转换运算符?

编写有缺陷的C ++代码是确定编译器错误的可靠方法:-) gcc中有超过1,000个已确认的未解决的错误。

是找到了模板化转化运算符should be,而"no match for 'operator*='"错误消息应改为"ambiguous overload for 'operator*='"

  

所以,在开启概念的情况下,我假设强制转换数字的唯一方法是将其转换为无符号整数类型 - 那么为什么编译器不能推断转换呢?

即使概念要求和编译器错误得到修复,模糊性仍然存在,特别是这四个:

main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned int)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned long long)
main.cpp:34:15: note: built-in candidate operator*=(unsigned int &, unsigned __int128)

这是因为每个可以想象的促销内置类型都有a lot个内置运算符,intlonglong long和{ {1}}都是整数类型。

出于这个原因,通常不是一个好主意来将转换模板化为内置类型。

解决方案1。制作转换运算符模板__int128并明确请求转换

explicit

解决方案2。只需实现 magnitude *= static_cast<uint32_t>(scale); // or magnitude *= static_cast<decltype(magnitude)>(scale); 类型的非模板化转换:

_number