编译时的质因数列表:专用化错误

List of prime factors at compile-time: specialization error

本文关键字：专用 错误 列表 质因数 编译      更新时间：2023-10-16

考虑以下代码（链接到IDEONE）：

#include <iostream>
#include <type_traits>
// List of factors
template<std::intmax_t ... Misc> 
struct factors { };
// Declaration
template<std::intmax_t ... Misc> 
struct factorization;
// Initial specialization
template<std::intmax_t Value> 
struct factorization<Value>
{
    typedef typename std::conditional<Value % 2 == 0,
            typename factorization<Value / 2, 2, 2>::type,
            typename factorization<Value / 2, 2 + 1>::type>::type type;
};
// Initial specialization when the value is not divisible by 2
template<std::intmax_t Value, std::intmax_t Divisor> 
struct factorization<Value, Divisor>
{
    typedef typename std::conditional<Value % Divisor == 0,
            typename factorization<Value / Divisor, Divisor, Divisor>::type,
            typename factorization<Value / Divisor, Divisor + 1>::type>::type type;
};
// Specialization after the first recusion step
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t Prime> 
struct factorization<Value, Divisor, Prime>
{
    typedef typename std::conditional<Value % Divisor == 0,
            typename factorization<Value / Divisor, Divisor, Divisor>::type,
            typename factorization<Value / Divisor, Divisor + 1>::type>::type type;
};
// Recursion specialization
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t Prime, std::intmax_t ... Primes> 
struct factorization<Value, Divisor, Prime, Primes...>
{
    typedef typename std::conditional<Value % Divisor == 0 && Divisor != Prime,
            typename factorization<Value / Divisor, Divisor, Divisor, Prime,
                    Primes...>::type,
            typename factorization<
                    Value % Divisor == 0 ? Value / Divisor : Value,
                    Divisor + (Value % Divisor != 0), Prime, Primes...>::type>::type type;
};
// Last recursion step
template<std::intmax_t Value, std::intmax_t ... Primes> 
struct factorization<Value, Value, Primes...>
{
    typedef typename factorization<1, Value, Value, Primes...>::type type;
};
// Finalize
template<std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization<1, Divisor, Primes...>
{
    typedef factors<Primes...> type;
};
// Main
int main() {
    typename factorization<18>::type x;
    return 0;
}

此代码应列出一个数字的质因数，不重复：例如，factorization<18>::type应等于factors<2, 3>

但它失败并显示以下错误：

prog.cpp: In instantiation of ‘struct factorization<4ll, 2ll, 2ll>’:
prog.cpp:36:79:   required from ‘struct factorization<9ll, 2ll, 2ll>’
prog.cpp:18:67:   required from ‘struct factorization<18ll>’
prog.cpp:67:28:   required from here
prog.cpp:36:79: error: ambiguous class template instantiation for ‘struct factorization<2ll, 2ll, 2ll>’
             typename factorization<Value / Divisor, Divisor + 1>::type>::type type;
                                                                               ^
prog.cpp:32:8: error: candidates are: struct factorization<Value, Divisor, Prime>
 struct factorization<Value, Divisor, Prime>
        ^
prog.cpp:41:8: error:                 struct factorization<Value, Divisor, Prime, Primes ...>
 struct factorization<Value, Divisor, Prime, Primes...>
        ^
prog.cpp:53:8: error:                 struct factorization<Value, Value, Primes ...>
 struct factorization<Value, Value, Primes...>
        ^
prog.cpp:36:79: error: invalid use of incomplete type ‘struct factorization<2ll, 2ll, 2ll>’
             typename factorization<Value / Divisor, Divisor + 1>::type>::type type;
                                                                               ^
prog.cpp:10:8: error: declaration of ‘struct factorization<2ll, 2ll, 2ll>’
 struct factorization;
        ^
prog.cpp: In function ‘int main()’:
prog.cpp:67:30: error: invalid combination of multiple type-specifiers
  typename factorization<18>::type x;
                              ^
prog.cpp:67:36: error: invalid type in declaration before ‘;’ token
  typename factorization<18>::type x;
                                    ^
prog.cpp:67:35: warning: unused variable ‘x’ [-Wunused-variable]
  typename factorization<18>::type x;
                                   ^
Compilation error   time: 0 memory: 0 signal:0
prog.cpp: In function ‘int main()’:
prog.cpp:67:30: error: ‘type’ in ‘struct factorization<18ll>’ does not name a type
  typename factorization<18>::type x;
                              ^
prog.cpp:67:36: error: invalid type in declaration before ‘;’ token
  typename factorization<18>::type x;
                                    ^
prog.cpp:67:35: warning: unused variable ‘x’ [-Wunused-variable]
  typename factorization<18>::type x;
                                   ^

如何解决这个问题？

这是一个工作版本。 请注意，使用 std::conditional 是一个坏主意，因为两个分支仍然需要扩展。 专业化效果要好得多。

// List of factors
template<std::intmax_t...> 
struct factors { };
// Declaration
template<bool no_remainder, std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization_remove_repeated;
template<bool no_remainder, std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization_check;
// wraps the remainder check to reduce code duplication
template<template<bool, std::intmax_t ...> class T, std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ...Primes> 
struct factorization_advance
{
    typedef typename T<(Value % Divisor) == 0, Value, Divisor, Primes...>::type type;
};
// end case
template<template<bool, std::intmax_t ...> class T, std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization_advance<T, 1, Divisor, Primes...>
{
    typedef factors<Primes...> type;
};
// No more repeats of Divisor, move to Divisor+1
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization_remove_repeated<false, Value, Divisor, Primes...>
{
    typedef typename factorization_advance<factorization_check, Value, Divisor + 1, Primes...>::type type;
};
// removed a repeat of Divisor, continue doing so, without adding to primes list
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes>
struct factorization_remove_repeated<true, Value, Divisor, Primes...>
{
    typedef typename factorization_advance<::factorization_remove_repeated, Value / Divisor, Divisor, Primes...>::type type;
};
// found that Divisor isn't a factor, move to Divisor+1
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes> 
struct factorization_check<false, Value, Divisor, Primes...>
{
    typedef typename factorization_advance<::factorization_check, Value, Divisor + 1, Primes...>::type type;
};
// Found first occurrence of a factor, add to primes list, remove repeats
template<std::intmax_t Value, std::intmax_t Divisor, std::intmax_t ... Primes>
struct factorization_check<true, Value, Divisor, Primes...>
{
    typedef typename factorization_advance<factorization_remove_repeated, Value / Divisor, Divisor, Primes..., Divisor>::type type;
};
// Convenience wrapper
template<std::intmax_t Value> 
struct factorization
{
    typedef typename factorization_advance<factorization_check, Value, 2>::type type;
};

  • 演示：http://rextester.com/NPGBU58803

  • 早期版本也可以工作，但可维护性较差的IMO：http://ideone.com/Ia9hnd

没有与

factorization<18>匹配的专业化（据我所知），除非您省略一些代码。你的每一个专业似乎都假设至少有两个参数被传递到因式分解中。

从您添加的评论来看，您似乎假设标记为"初始专业化"的专业化将捕获您只传入单个数字的情况。这不是真的。尝试运行factorization<18, 2>以查看其实际效果。如果这确实是您想要的，那么要么省略您的专业化中的 2 并保持身体不变，要么添加一个最终专业化，该专业仅匹配单个数字并将 type 设置为 typename factorization<N, 2>::type .

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