c++ - std::abs(std::complex) 太慢

Question

为什么在大型复杂数组上运行 std::abs 比使用 sqrt 和 norm 慢大约 8 倍？

#include <ctime>
#include <cmath>
#include <vector>
#include <complex>
#include <iostream>
using namespace std;

int main()
{
    typedef complex<double> compd;

    vector<compd> arr(2e7);
    for (compd& c : arr)
    {
        c.real(rand());
        c.imag(rand());
    }

    double sm = 0;
    clock_t tm = clock();
    for (const compd& c : arr)
    {
        sm += abs(c);
    }
    cout << sm << ' ' << clock() - tm << endl; // 5.01554e+011 - 1640 ms

    sm = 0;
    tm = clock();
    for (const compd& c : arr)
    {
        sm += sqrt(norm(c));
    }
    cout << sm << ' ' << clock() - tm << endl; // 5.01554e+011 - 154

    sm = 0;
    tm = clock();
    for (const compd& c : arr)
    {
        sm += hypot(c.real(), c.imag());
    }
    cout << sm << ' ' << clock() - tm << endl; // 5.01554e+011 - 221
}

Answer 1

我认为两者在严格意义上不能被视为相同。

来自 cppreference on std::abs(std::complex) :

Errors and special cases are handled as if the function is implemented as std::hypot(std::real(z), std::imag(z))

同样来自 cppreference on std::norm(std::complex) :

The norm calculated by this function is also known as field norm or absolute square.

The Euclidean norm of a complex number is provided by std::abs, which is more costly to compute. In some situations, it may be replaced by std::norm, for example, if abs(z1) > abs(z2) then norm(z1) > norm(z2).

简而言之，存在从每个函数获得不同结果的情况。其中一些可以在 std::hypot 中找到.注释中还提到了以下内容:

std::hypot(x, y) is equivalent to std::abs(std::complex<double>(x,y))

一般来说，结果的准确性可能会有所不同(由于通常的浮点困惑)，函数的设计似乎是为了尽可能准确。