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在编译时将十进制转换为十六进制

[英]Converting decimal to hexadecimal at compile time

I'm trying to create a program that can do conversions from a decimal number to a hexidecimal number at compile time. 我正在尝试创建一个程序,该程序可以在编译时从十进制数转换为十六进制数。 Unfortunately I'm new to metaprogramming and I can't figure out a way to do this at compile time. 不幸的是,我是元编程的新手,在编译时我找不到一种方法。 Any ideas how I can approach this? 有什么想法可以解决这个问题吗?

How can I represent such a number without violating the principles of template metaprogramming. 如何在不违反模板元编程原理的情况下表示这样的数字。

#include <algorithm>        // std::reverse
#include <iterator>         // std::begin, std::end
#include <iostream>
#include <tuple>
#include <type_traits>      // std::integral_constant

namespace detail {
    using std::tuple;
    using std::tuple_cat;
    using std::integral_constant;

    template< unsigned x, int n_digits >
    struct Hex_tuple_
    {
        using Type = decltype( tuple_cat(
            typename Hex_tuple_< x & ((1 << 2*n_digits) - 1), n_digits/2 >::Type(),
            typename Hex_tuple_< (x >> 2*n_digits), n_digits/2 >::Type()
            ) );
    };

    template< unsigned x >
    struct Hex_tuple_< x, 1 >
    {
        using Type = tuple< integral_constant< int, x > >;
    };

}  // namespace detail

template< unsigned x >
using Hex_ = typename detail::Hex_tuple_< x, 2*sizeof( x ) >::Type;

template< int... digit >
auto operator<<(
    std::ostream& stream,
    const std::tuple< std::integral_constant< int, digit >... >&
    )
    -> std::ostream&
{
    int digits[] = { digit... };
    std::reverse( std::begin( digits ), std::end( digits ) );
    for( int x : digits )
    {
        stream << "0123456789ABCDEF"[x];
    }
    return stream;
}

auto main() -> int
{
    std::cout << Hex_<123456>() << "\n";
}

The general problem of constructing the base B numeral that represents a given unsigned integer at compiletime is quite interesting. 构建在编译时表示给定无符号整数的基数B的一般问题非常有趣。 It can be solved concisely by an application of the valuable technique of building a compiletime std::array , which goes back at least to this answer 可以通过应用构建编译时std::array的宝贵技术来简洁地解决此问题 ,至少可以回溯到此答案

The function template digits as defined below returns a compiletime null-terminated std::array of char that contains the base Base numeral representing the integer N , for Base in the reasonable range [2,36] (digits '0-9A-Z') 如下定义的函数模板digits将返回一个以char表示的编译时以空值终止的std::array ,其中char包含表示整数N的基本Base数字,对于在合理范围[2,36]中的Base (数字'0-9A-Z' )

#include <type_traits>
#include <array>

// Get the number of base `base` digits in `n`
constexpr std::size_t base_digits(unsigned long n, unsigned base) 
{
    return (n / base) == 0 ? 1 : 1 + base_digits(n / base,base);
}

// Convert the remainder `r` of division by some base to a based digit
constexpr char based_digit(unsigned r)
{
    return r > 36 ? '!' : r < 10 ? '0' + r : 'A' + r - 10;
}

template <unsigned long N, unsigned Base, 
    std::size_t Len = base_digits(N,Base) + 1, unsigned ...Rems>
constexpr std::array<char, Len> 
digits(std::enable_if_t<(Base > 1 && Base <= 36 && N < Base)> * = nullptr)
{
    return std::array<char,Len>{{based_digit(N), Rems...,0}};
}

template <unsigned long N, unsigned Base, 
    std::size_t Len = base_digits(N,Base) + 1, unsigned ...Rems>
constexpr std::array<char,Len> 
digits(std::enable_if_t<(Base > 1 && Base <= 36 && N >= Base)> * = nullptr)
{
    return digits<N / Base, Base, Len, based_digit(N % Base), Rems...>();  
}

Prepend that to this test: 在此测试之前添加:

constexpr auto decimal = digits<9876543210,10>();
constexpr auto hex = digits<9876543210,16>();
constexpr auto octal = digits<123456789,8>();
constexpr auto binary = digits<123456789,2>();
constexpr auto b_36 = digits<36 + 35,36>();

static_assert(decimal.size() == 11,"");

#include <iostream>
#include <cassert>

int main() {
    char * pend;    
    assert(std::strtoul(decimal.data(),&pend,10) == 9876543210); 
    std::cout << decimal.data() << std::endl;
    assert(std::strtoul(hex.data(),&pend,16) == 9876543210);
    std::cout << hex.data() << std::endl;
    assert(std::strtoul(octal.data(),&pend,8) == 123456789);
    std::cout << octal.data() << std::endl;
    assert(std::strtoul(binary.data(),&pend,2) == 123456789);
    std::cout << binary.data() << std::endl;
    assert(std::strtoul(b_36.data(),&pend,36) == 36 + 35);
    std::cout << b_36.data() << std::endl;
    return 0;
}

Output is: 输出为:

9876543210
24CB016EA
726746425
111010110111100110100010101
1Z

(gcc 4.9.2/clang 3.5.1, -std=c++14. Trivially adaptable for C++11) (gcc 4.9.2 / clang 3.5.1,-std = c ++ 14。适用于C ++ 11)

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