将整数模板参数值映射到原始类型 [英] Mapping an integral template parameter value onto a primitive type

问题描述

我想将数字映射到类型。对于此示例，我将创建一个将sizeof（）结果映射到带符号原始类型的函数。

I want to map a number to a type. For this example I'll make a function that maps a sizeof() result onto a signed primitive type.

我想知道是否有更好的方法来执行我的操作下面是在现代C ++中执行的操作，它采用模板值并将其转换为类型。现在，这可以将大小转换为已知类型，但是我似乎在标准库中找不到能满足我需要的任何东西。我错过了什么吗？

I am wondering if there is a better way of doing what I did below in modern C++, which is to take a templated value and convert it into a type. Right now this works in converting a size to a known type, but I can't seem to find anything in the standard library that does what I want. Have I missed something?

如果没有，是否有更好的方法来执行此操作或清除此代码？例如，如果将来以某种方式我们最终拥有128位类型，则将不支持该类型。

If not, is there a better way of doing this or cleaning up this code? For example if somehow in the future we end up having 128 bit types, this wouldn't support that.

#include <iostream>
#include <type_traits>

template <size_t S>
static constexpr auto sizeToType() {
    static_assert(S == 1 or S == 2 or S == 4 or S == 8, "Bad type size");

    if constexpr (S == 1)
        return int8_t{};
    else if constexpr (S == 2)
        return int16_t{};
    else if constexpr (S == 4)
        return int32_t{}; 
    else
        return int64_t{};
}

int main() {
    using MY_TYPE = decltype(sizeToType<2>());

    MY_TYPE myType = MY_TYPE(0xFFFFFFFFFFFFFFFEUL);

    std::cout << sizeof(MY_TYPE) << " bytes" << std::endl;
    std::cout << "MY_TYPE(0xFFFFFFFFFFFFFFFEUL) = " << myType << std::endl;
}

输出（预期）：

2 bytes
MY_TYPE(0xFFFFFFFFFFFFFFFEUL) = -2

推荐答案

为此，我不会使用C ++ 17 constexpr ，而是使用模板专用化，

I would not use C++17 if constexpr for this, instead I'd use template specialization, as it looks more declarative to me.

以下内容：

template<size_t S> struct SizeToType {static_assert(S != S, "Wrong size"); };

template<> struct SizeToType<1> { using type = uint8_t; };
template<> struct SizeToType<2> { using type = uint16_t; };
template<> struct SizeToType<4> { using type = uint32_t; };
template<> struct SizeToType<8> { using type = uint64_t; };

template<size_t S>
using SizeToToTypeT = typename SizeToType<S>::type;

添加更多类型只会在此处添加更多专长（单线）。

Adding more types would be just adding more specializations (one-liners) here.

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