将二维数组元素复制到另一个具有其他大小的二维数组中

哦，多么好的编程难题啊。我的解决方案是将两个数组都展平并复制。

template <typename T>
static constexpr T* begin(T& value) noexcept
{
  return &value;
}
template <typename T, ::std::size_t N>
static constexpr typename ::std::remove_all_extents<T>::type*
begin(T (&array)[N]) noexcept
{
  return begin(*array);
}
template <typename T>
static constexpr T* end(T& value) noexcept
{
  return &value + 1;
}
template <typename T, ::std::size_t N>
static constexpr typename ::std::remove_all_extents<T>::type*
end(T (&array)[N]) noexcept
{
  return end(array[N - 1]);
}
int a[4][4];
int b[2][8];
::std::copy(begin(a), end(a), begin(b));

为什么不从输入矩阵中创建一个临时线性数组，然后用它来填充输出矩阵：

void changearr ( int** ppm, int old_row, int old_col, int new_row, int new_col )
{
    int* temp_linear = new int[old_row * old_col];
    int k = 0;
    for ( int i = 0; i < old_row; i++ )
    {
        for ( int j = 0; j < old_col; j++ )
        {
            temp_linear[k++] = ppm[i][j];
        }
    }
    int** temp = new int* [new_row];
    for ( int i = 0; i < new_row; i++ )
    {
        temp[i] = new int[new_col];
    }
    k = 0;
    for ( int i = 0; i < new_row; i++ )
    {
        for ( int j = 0; j < new_col; j++ )
        {
            if ( k < old_row * old_col )
            {
                temp[i][j] = temp_linear[k++];
            }
            else
            {
                temp[i][j] = 0;
            }
        }
    }
}

您可以使用简单的临时容器：

#include <iostream>
#include <deque>
#include <array>
int main()
{
    std::array<std::array<int,4>,4> first {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}; //4x4
    std::array<std::array<int,8>,2> second; //2x8
    std::deque<int> temp; //temporary container
    for(auto x : first)
        for(auto y : x)
            temp.push_back(y); //push everything from first to deque
    for(auto& x : second)
        for(auto& y : x)
        {
            y = temp.front(); //move from deque to second and pop()
            temp.pop_front();
        }
}

2D数组是而不是指针对指针是一样的！作为一个2D数组，只需一行接一行地存储行，您只需复制元素：

#include <iostream>
void trans(int* orig, int rows1, int cols1, int *resul, int rows2, int cols2) {
    int tot1 = rows1 * cols1;
    int tot2 = rows2 * cols2;
    int tot = tot1;
    if (tot2 < tot) tot = tot2;
    // copy the smallest size from both arrays
    for(int i=0; i<tot; i++) {
        resul[i] = orig[i];
    }
    // eventually add 0 to fill resul array
    for(int i=tot; i<tot2; i++) {
        resul[i] = 0;
    }
}
int main() 
{ 
    int orig[4][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6};
    // control initial array
    for (int i=0; i<4; i++) {
        for (int j=0; j<4; j++) {
            std::cout << orig[i][j];
        }
        std::cout << std::endl;
    }
    int resul[2][8];
    // pass arrays as 1D array pointers!
    trans(reinterpret_cast<int *>(&orig), 4, 4, reinterpret_cast<int *>(&resul), 2, 8);
    // control converted array
    for (int i=0; i<2; i++) {
        for (int j=0; j<8; j++) {
            std::cout << resul[i][j];
        }
        std::cout << std::endl;
    }
    return 0;
}

如预期：

1234
5678
9012
3456
12345678
90123456

如您所知，C/C++n维数组在内存上具有线性表示。因此，如果你将打开这种表示，你会得到以下好处：

std::vector<int> v1{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
for(size_t r = 0; r < 4; r++) 
for(size_t c = 0; c < 4; c++) 
    std::cout << v1[r*4+c];
std::vector<int> v2(v1);
for(size_t r = 0; r < 2; r++) 
for(size_t c = 0; c < 8; c++) 
    std::cout << v2[r*8+c];

这里有一个非常简单的方法：

void
   changearr(
     int **in,int in_size1,int in_size2,
     int **out,int out_size1,int out_size2
   )
{
  int in_n = in_size1*in_size2;
  int out_n = out_size1*out_size2;
  int n = min(in_n,out_n);
  int i = 0;
  for (; i!=n; ++i) {
    out[i/out_size2][i%out_size2] = in[i/in_size2][i%in_size2];
  }
  for (; i!=out_n; ++i) {
    out[i/out_size2][i%out_size2] = 0;
  }
}

这个想法只是线性地迭代索引，并根据需要计算输入和输出数组的对应索引。