std::sqrt(std::valarray)

来自cppreference.com
< cpp‎ | numeric‎ | valarray
 
 
 
 
在标头 <valarray> 定义
template< class T >
valarray<T> sqrt( const valarray<T>& va );

va 中每个元素计算元素值的平方根。

参数

va - 要应用操作到的值数组

返回值

含有 va 中值的平方根的值数组。

注解

用无限定函数 (sqrt) 进行计算。若该函数不可用,则会由于实参依赖查找而使用 std::sqrt

函数可以实现为拥有不同于 std::valarray 的返回类型。此时替换它的类型拥有下列属性:

可能的实现

template<class T>
valarray<T> sqrt(const valarray<T>& va)
{
    valarray<T> other = va;
    for (T& i : other)
        i = sqrt(i);
 
    return other; // 可以返回代理对象
}

示例

一次寻找几个三次方程的所有三个根(其中两个为复共轭)。

#include <cassert>
#include <complex>
#include <cstddef>
#include <iostream>
#include <numbers>
#include <valarray>
 
using CD = std::complex<double>;
using VA = std::valarray<CD>;
 
// 返回给定复数 x 的所有 n 个复根
VA root(CD x, unsigned n)
{
    const double mag = std::pow(std::abs(x), 1.0 / n);
    const double step = 2.0 * std::numbers::pi / n;
    double phase = std::arg(x) / n;
    VA v(n);
    for (std::size_t i{}; i != n; ++i, phase += step)
        v[i] = std::polar(mag, phase);
    return v;
}
 
// 返回 v 中各元素的 n 复根;输出 valarray 中首先包含 v[0] 的所有 n 个根,
// 然后是 v[1] 的所有 n 个根,等等。
VA root(VA v, unsigned n)
{
    VA o(v.size() * n);
    VA t(n);
    for (std::size_t i = 0; i != v.size(); ++i)
    {
        t = root(v[i], n);
        for (unsigned j = 0; j != n; ++j)
            o[n * i + j] = t[j];
    }
    return o;
}
 
// 容许给定舍入误差的浮点数比较器
inline bool is_equ(CD x, CD y, double tolerance = 0.000'000'001)
{
    return std::abs(std::abs(x) - std::abs(y)) < tolerance;
}
 
int main()
{
    // 多项式 x³ + p·x + q 的输入系数
    const VA p{1, 2, 3, 4, 5, 6, 7, 8};
    const VA q{1, 2, 3, 4, 5, 6, 7, 8};
 
    // 求解器
    const VA d = std::sqrt(std::pow(q / 2, 2) + std::pow(p / 3, 3));
    const VA u = root(-q / 2 + d, 3);
    const VA n = root(-q / 2 - d, 3);
 
    // 为根分配内存:3 * 三次多项式个数
    VA x[3];
    for (std::size_t t = 0; t != 3; ++t)
        x[t].resize(p.size());
 
    auto is_proper_root = [](CD a, CD b, CD p) { return is_equ(a * b + p / 3.0, 0.0); };
 
    // 从(每个多项式)产生的 9 个根中筛出 6 个,仅保留 3 个根
    for (std::size_t i = 0; i != p.size(); ++i)
        for (std::size_t j = 0, r = 0; j != 3; ++j)
            for (std::size_t k = 0; k != 3; ++k)
                if (is_proper_root(u[3 * i + j], n[3 * i + k], p[i]))
                    x[r++][i] = u[3 * i + j] + n[3 * i + k];
 
    std::cout << "Depressed cubic equation:   Root 1: \t\t Root 2: \t\t Root 3:\n";
    for (std::size_t i = 0; i != p.size(); ++i)
    {
        std::cout << "x³ + " << p[i] << "·x + " << q[i] << " = 0  "
                  << std::fixed << x[0][i] << "  " << x[1][i] << "  " << x[2][i]
                  << std::defaultfloat << '\n';
 
        assert(is_equ(std::pow(x[0][i], 3) + x[0][i] * p[i] + q[i], 0.0));
        assert(is_equ(std::pow(x[1][i], 3) + x[1][i] * p[i] + q[i], 0.0));
        assert(is_equ(std::pow(x[2][i], 3) + x[2][i] * p[i] + q[i], 0.0));
    }
}

输出:

Depressed cubic equation:   Root 1:              Root 2:                 Root 3:
x³ + (1,0)·x + (1,0) = 0  (-0.682328,0.000000)  (0.341164,1.161541)  (0.341164,-1.161541)
x³ + (2,0)·x + (2,0) = 0  (-0.770917,0.000000)  (0.385458,1.563885)  (0.385458,-1.563885)
x³ + (3,0)·x + (3,0) = 0  (-0.817732,0.000000)  (0.408866,1.871233)  (0.408866,-1.871233)
x³ + (4,0)·x + (4,0) = 0  (-0.847708,0.000000)  (0.423854,2.130483)  (0.423854,-2.130483)
x³ + (5,0)·x + (5,0) = 0  (-0.868830,0.000000)  (0.434415,2.359269)  (0.434415,-2.359269)
x³ + (6,0)·x + (6,0) = 0  (-0.884622,0.000000)  (0.442311,2.566499)  (0.442311,-2.566499)
x³ + (7,0)·x + (7,0) = 0  (-0.896922,0.000000)  (0.448461,2.757418)  (0.448461,-2.757418)
x³ + (8,0)·x + (8,0) = 0  (-0.906795,0.000000)  (0.453398,2.935423)  (0.453398,-2.935423)

参阅

应用函数 std::pow 到二个 valarray 或一个 valarray 与一个值
(函数模板)
(C++11)(C++11)
计算平方根(x
(函数)
右半平面范围中的复平方根
(函数模板)