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c++ - 模板推导与隐式用户定义转换运算符

转载 作者:可可西里 更新时间:2023-11-01 17:59:07 25 4
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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 intlonglong long__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/

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