<分区>
我正在比较这两个函数:
double polynomials(const vector<double> & coeffs, double x) {
double sum = 0.0;
double factor = 1.0;
for (double coeff: coeffs) {
sum += coeff * factor;
factor *= x;
}
return sum;
}
和
double algorithm_polynomials(const vector<double> & coeffs, double x) {
return reduce(execution::seq, cbegin(coeffs), end(coeffs), 0, [factor=1.0, x](double sum, double coeff) mutable {
double curr_sum = sum + coeff * factor;
factor *= x;
return curr_sum;
});
}
对于整数系数和整数x，这两个函数的结果是等价的。但对于实际值，结果有所不同。第一个函数给出的是正确的，而第二个函数给出的总是整数(不一定接近真实答案)。
我的猜测是每个中间值在求和之前都被转换为整数，但我不明白为什么会发生这种情况，或者我该如何解决它。
要运行此代码，请使用
#include <numeric>
#include <vector>
#include <algorithm>
#include <execution>
using namespace std;
需要带有 -std=c++17 标志的 gcc 9 或更高版本以及从 2019 年到 tbb 的链接。
编辑:感谢大家指出我忽略的初始值类型。
我感到有点惭愧，既因为没有看到它，也因为 reduce 函数使用不当(由于可变的 lambda 而无法将其并行化)。
为了救赎自己，我在这里添加了一种使其并行的方法...
class counter: public std::iterator<
std::random_access_iterator_tag, // iterator_category
size_t, // value_type
size_t, // difference_type
const size_t*, // pointer
size_t // reference
>{
size_t num = 0;
public:
explicit counter(size_t _num) : num(_num) {}
counter& operator++() {num += 1; return *this;}
counter operator++(int) {counter retval = *this; ++(*this); return retval;}
bool operator==(counter other) const {return num == other.num;}
bool operator!=(counter other) const {return !(*this == other);}
counter& operator+=(size_t i) { num += i; return *this; }
counter& operator-=(size_t i) { num -= i; return *this; }
counter operator +(counter &other) const { return counter(num + other.num);}
counter operator -(counter &other) const { return counter(num - other.num); }
counter operator +(size_t i) const { return counter(num + i); }
counter operator -(size_t i) const {return counter(num - i); }
reference operator*() const {return num;}
};
double better_algorithm_polinomials(const vector<double> & coeffs, double x) {
//this has the advantage of being easily parallelized
return transform_reduce(execution::par, cbegin(coeffs), end(coeffs), counter(0), 0.0, plus{}, [x](double coeff, size_t index) { return coeff * pow<double>(x, index); });
}