想要创建一个2D指针，在C ++中存储Nx3值

Question

我已经在下面编写了代码，并且希望使用if-else条件使用它存储一些值。 为了存储值，我创建了两个2-D指针* p和* q。 现在代码似乎可以正常运行，但是我无法打印出存储在指针p和q中的值。 你能帮我吗？ 我究竟做错了什么”？

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>

int main()
{
    int i, j, N;
    double xtmp, ytmp, ztmp, delx, dely, delz, fpair, cutsq;
    double rsq, r2inv, r6inv, forcelj; 
    double firstval[2][3], secondval[2][3];
    double (*p)[2][3] = 0; 
    double (*q)[2][3] = 0;
    int f[2][3];
    p =& firstval;
    q =& secondval;

    double x[2][3] =
    {
        {0.1,0.8,0.12},
        {0.5,0.9,0.13}
    };

    N = 2;
    cutsq = 1.0;

    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            xtmp = x[i][0];
            ytmp = x[i][1];
            ztmp = x[i][2]; 

            delx = xtmp - x[j][0];
            dely = ytmp - x[j][1];
            delz = ztmp - x[j][2];
            rsq = delx * delx + dely * dely + delz * delz;
            if (rsq < cutsq) 
            {
                r2inv = 1.0 / rsq;
                r6inv = r2inv * r2inv * r2inv;
                forcelj = r6inv * 1;
                fpair = forcelj * r2inv;

                f[i][0] += delx * fpair;
                f[i][1] += dely * fpair;
                f[i][2] += delz * fpair;
                f[j][0] -= delx * fpair;
                f[j][1] -= dely * fpair;
                f[j][2] -= delz * fpair;

                if (i != j)
                {
                    (*p)[i][0] += delx * fpair;
                    (*p)[i][1] += dely * fpair;
                    (*p)[i][2] += delz * fpair;
                    (*p)[j][0] -= delx * fpair;
                    (*p)[j][1] -= dely * fpair;
                    (*p)[j][2] -= delz * fpair;
                 } 
                 else 
                 {
                     (*q)[i][0] += delx * fpair;
                     (*q)[i][1] += dely * fpair;
                     (*q)[i][2] += delz * fpair;
                     (*q)[j][0] -= delx * fpair;
                     (*q)[j][1] -= dely * fpair;
                     (*q)[j][2] -= delz * fpair;
                 }

                 std::cout << firstval << "\n";
            }
        }
    }         
}

Answer 1

我毫不怀疑有一种更聪明的方法，但是...
你可以写
a）“知道”如何流double [N]的模板，并且
b）知道（基于第一个）如何流double [M] [N]的模板

template <typename T, std::size_t N>
std::ostream& operator<< (std::ostream &o, const T(&a)[N]) {
    o << '[' << a[0];  // iirc zero-sized arrays are not allowed ....better look it up though.
    for (std::size_t i = 1; i < N; ++i) {
        o << ',' << a[i];
    }
    o << ']';
    return o;
}

template <typename T, std::size_t M, std::size_t N>
std::ostream& operator<< (std::ostream &o, const T(&a)[M][N]) {
    o << '{' << a[0];
    for (std::size_t i = 1; i < M; ++i) {
        o << ',' << a[i];
    }
    o << '}';
    return o;
}

然后您可以执行std::cout << firstval << std::endl;

---

编辑：当然，还有一个更通用的解决方案

template <typename T> void print_array(const T & a, std::size_t level=0);
template <typename T>
void print_array_impl(const T & a, std::size_t level, std::true_type)
{
    std::cout << std::endl;
    for (auto & x : a) print_array(x, level+1);
    fill_n(std::ostream_iterator<char>(std::cout), level, ' ');
}

template <typename T>
void print_array_impl(const T & a, std::size_t level, std::false_type)
{
    for (auto & x : a) std::cout << x << ',';
}

template <typename T>
void print_array(const T & a, std::size_t level)
{
    fill_n(std::ostream_iterator<char>(std::cout), level, ' ');
    std::cout << '[';
    print_array_impl(a, level, std::is_array<typename std::remove_extent<T>::type>());
    std::cout << ']' << std::endl;

}

.oO（我需要重新整理一下c ++;看起来还是不太优雅....）