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c++ - 具有数组构造函数方法的 constexpr 类

转载 作者:搜寻专家 更新时间:2023-10-31 00:39:11 24 4
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我正在尝试为多元多项式编写一个编译时类(即像 P(X,Y,Z) = X^2 + XYZ + YZ,不要太担心这里的数学):

template<int DIM, int DEGREE> class Polynomial {
public:
constexpr Polynomial(std::array<double,nbOfCoeffs(DIM,DEGREE)> arr): coeffs(arr) {}


constexpr double eval(std::array<double,DIM> x);
constexpr operator+,-,*,/ ...
private:
std::array<double,nbOfCoeffs(DIM,DEGREE)> coeffs; //don't worry about "nbOfCoeffs" : it is constexpr and computes at compile time the right number of coefficients.
}

int main () {
Polynomial<2,2> P({{1.,1.,1.,1.,1.,1.}}); // P(X,Y) = X^2+XY+Y^2+X+Y+1

double x = P.eval(1.);
auto P2 = P*P;
}

到目前为止,实现这个没有什么大问题。但是,请注意我的 ctor 可能有点麻烦:如何初始化三阶三元多项式 P(X,Y,Z) = XYZ?我会有类似的东西

多项式<3,3> P({{0.,0.,0.,0.,0.,1.,0.,0.,0.,0.}});

唯一非零单项式的位置取决于我存储它的位置。如果我能写下就好了:

  Polynomial<3,3> P("XYZ"); 

Polynomial<4,3> P("X1X2X3 + X4^2"); //more general

想法是创建某种小型 DST 来处理多项式的字符串表示。

但是,如果我这样做,一旦字符串被解析,我就不知道如何为我的存储数组的元素赋值:所有这些都必须在 ctor 的主体保持为空的情况下完成(因为我希望它是构造函数)。你怎么看呢 ?是否可以 ?我是否应该将我的数组更改为某种重复结构(因为在这种情况下,我认为它会变得非常非常复杂)

最佳答案

如何实现 Luc Danton 的示例的做法:

#include <cstddef>
#include <iostream>

namespace polynomials
{
// it's possible to store the exponent as data member instead
template < std::size_t t_id, std::size_t t_exponent = 1 >
struct monomial
{
static constexpr std::size_t id = t_id;
static constexpr std::size_t exponent = t_exponent;

// it's not possible to store the coefficient
// as non-type template parameter (floating-point..)
double coefficient;

explicit constexpr monomial(double p_coefficient = 1.0)
: coefficient{ p_coefficient }
{}

void print() const
{
std::cout << coefficient << "X" << t_id << "^" << t_exponent;
}
};

// create the monomial objects (like std::placeholders::_1)
constexpr monomial<0> X;
constexpr monomial<1> Y;
constexpr monomial<2> Z;

constexpr monomial<4> X0;
constexpr monomial<5> X1;
// ... can use macros to produce a lot of them..

// multiply an arithmetic type (double, int, ..) with a monomial
template < typename T_Arithmetic,
std::size_t t_id, std::size_t t_exponent0 >
constexpr auto operator*(T_Arithmetic c, monomial < t_id, t_exponent0 > m0)
-> monomial < t_id, t_exponent0 >
{
return monomial < t_id, t_exponent0 >{c * m0.coefficient};
}
// the other way 'round
template < typename T_Arithmetic,
std::size_t t_id, std::size_t t_exponent0 >
constexpr auto operator*(monomial < t_id, t_exponent0 > m0, T_Arithmetic c)
-> monomial < t_id, t_exponent0 >
{
return c * m0;
}

// multiply two monomials with the same id
template < std::size_t t_id,
std::size_t t_exponent0, std::size_t t_exponent1 >
constexpr auto operator*(monomial < t_id, t_exponent0 > m0,
monomial < t_id, t_exponent1 > m1)
-> monomial < t_id, t_exponent0 + t_exponent1 >
{
return monomial<t_id, t_exponent0 + t_exponent1>
{m0.coefficient * m1.coefficient};
}


// storage type for multiple different monomials
template < typename... T_Monomials >
struct polynomial
{
void print() const
{}
};
template < typename T_Monomial, typename... TT_Monomials >
struct polynomial < T_Monomial, TT_Monomials... >
: public polynomial < TT_Monomials... >
{
using base = polynomial < TT_Monomials... >;

T_Monomial m;
constexpr polynomial(T_Monomial p, TT_Monomials... pp)
: base(pp...)
, m{p}
{}

void print() const
{
m.print();
std::cout << "*";
base::print();
}
};

// multiply two monomials to get a polynomial
template < std::size_t t_id0, std::size_t t_id1,
std::size_t t_exponent0, std::size_t t_exponent1 >
constexpr auto operator*( monomial < t_id0, t_exponent0 > m0,
monomial < t_id1, t_exponent1 > m1)
-> polynomial < monomial<t_id0, t_exponent0>,
monomial<t_id1, t_exponent1> >
{
return {m0, m1};
}

// still to do (and more complicated):
// - multiply two polynomials
// - multiply a polynomial and a monomial
// - addition, subtraction, division (?) etc.
}

使用示例:

int main()
{
using namespace polynomials;

auto p0 = 1.25*X*X;
p0.print();
std::cout << std::endl;

auto p1 = p0 * 5*Y;
p1.print();
std::cout << std::endl;
}

关于c++ - 具有数组构造函数方法的 constexpr 类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16638724/

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