我试图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: In function ‘void changeScale(uint32_t&, Number)’:

main.cpp:40:15: error: no match for ‘operator*=’ (operand types are ‘uint32_t {aka unsigned int}’ and ‘Number’)

magnitude *= scale;

注意注释掉的行 - 这行有效:

//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]

我按照删除概念的建议使用 Clang 并查看了它的错误消息。我得到以下信息(被截断但足够了):

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，打开这些概念后，我假设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足以推断转换？

最佳答案

requires 概念表达式的工作方式类似于 SFINAE，它只检查表达式是否有效，但不对其求值。

要使概念实际上限制 T 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>;

这会解决您的问题吗？不幸的是，请继续阅读...

Why can't the templated conversion operator be automatically deduced?

编写错误的 C++ 代码肯定会遇到编译器错误:-) gcc 中有超过 1,000 个已确认的未解决错误。

是模板化转换运算符should be找到了，并且 "no match for 'operator*='" 错误消息应该改为 "ambiguous overload for 'operator*='"。

So, having the concepts turned on I assume that the only way to cast a Number is to do it to an unsigned integral type - then why is it insufficient for the compiler to deduce the conversion?

即使修复了概念要求和编译器错误，歧义仍然存在，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)

那是因为有a lot每个可以想到的提升内置类型的内置运算符，以及 int、long、long long 和 __int128 都是整数类型。

因此，将转换模板化为内置类型通常不是一个好主意。

解决方案 1. 使转换运算符模板显式并显式请求转换

    magnitude *= static_cast<uint32_t>(scale);
    // or
    magnitude *= static_cast<decltype(magnitude)>(scale);

解决方案 2. 只需实现对 _number 类型的非模板转换:

struct Number
{
    using NumberType = uint32_t;
    operator NumberType () const
    {
        return this->_number;
    }
    NumberType _number;
};

关于c++ - 模板推导与隐式用户定义转换运算符，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49646290/