将二维数组元素复制到另一个具有另一个大小的二维数组

Question

我试图将二维数组复制到另一个大小的另一个数组.例如:第一个包含4行和4列的数组:

1 2 3 4
5 6 7 8
9 0 1 2
3 4 5 6

第二个数组有2行和8列:

1 2 3 4 5 6 7 8
9 0 1 2 3 4 5 6

如果新数组中的元素多于第一个元素,则函数将用0填充它

这是我做的功能,但索引的问题.如何以正确的方式写出来？

void changearr(int **ppm, int size1, int size2, int size_1, int size_2)
{
    int **temp = new int*[size_1];
    for (int i = 0; i < size_1; i++)
        temp[i] = new int[size_2];  
    int z = 0;
    for (int i = 0; i < size_1; i++, z++)
    {
        for (int j = 0, k = 0; j < size_2; j++, k++)
        {
            if (i < size_1 || j < size_2)
            {
                temp[i][j] = ppm[z][k];
            }
            else
                temp[i][j] = 0
        }
    }

Answer 1

哦,这是一个很棒的编程难题.我的解决方案是将两个数组展平并复制它们.

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));

Answer 2

为什么不从输入矩阵创建一个临时线性数组,然后使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;
            }
        }
    }
}

Answer 3

你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();
        }
}