如何在 C/C++ 中获取多维数组的列？

Question

int matrix[9][9],*p;
p=matrix[0]; 

this works and gives first row of matrix , but how to get first column of matrix I've tried p=matrix[][0]; ? Also I don't understand why below code gets compiler error?

int matrix[9][9],p[9];  // it looks really ugly, byt why it doesn't work ?
p=matrix[0];            // compiler gives "invalid array assigment"

is it because multidimensional arrays are arrays of arrays - and we should interpret matrix[i][j] as j-th element of i-th nested array?

Answer 1

In C/C++, multidimensional arrays are actually stored as one dimensional arrays (in the memory). Your 2D matrix is stored as a one dimensional array with row-first ordering. That is why getting a column out of it is not easy, and not provided by default. There is no contiguous array in the memory that you can get a pointer to which represents a column of a multidimensional array. See below:

When you do p=matrix[0] , you are just getting the pointer to the first element matrix[0][0] , and that makes you think that you got the pointer to first row. Actually, it is a pointer to the entire contiguous array that holds matrix , as follows:

matrix[0][0]
matrix[0][1]
matrix[0][2]
.
.
matrix[1][0]
matrix[1][1]
matrix[1][2]
.
.
matrix[8][0]
matrix[8][1]
matrix[8][2]
.
.
matrix[8][8]

As seen above, the elements of any given column are separated by other elements in the corresponding rows.

So, as a side note, with pointer p , you can walk through the entire 81 elements of your matrix if you wanted to.

Answer 2

You can get the first column using a loop like

for(int i = 0; i < 9; i++)
{
    printf("Element %d: %d", i, matrix[i][0]);
}

I think the assignment doesn't work properly because you're trying to assign something's that's not an address to a pointer.

(Sorry this is c code)

Answer 3

There is no difference between specifying matrix[81] or matrix[9][9]

matrix[r][c] simply means the same as matrix[9*r+c]

There are other containers better suited fort multidimensional arrays like boost::multi_array

http://www.boost.org/doc/libs/1_53_0/libs/multi_array/doc/index.html

Think of the bare array just like allocating a contiguous piece of memory. You, the programmer then has to handle this piece of memory yourself. The bare name of the array, eg matrix is a pointer to the first element of this allocated piece of memory. Then *(matrix+1) is the same as matrix[0][1] or matrix[1] .

Answer 4

p 是一个 int 数组，matrix[0] 是一个指针..

Answer 5

matrix本身是最接近数组列的东西，因为(matrix + 1)[0][0]与matrix[1][0] 。

Answer 6

If you want your matrix to contiguous locations, declare it as a one dimensional array and perform the row and column calculations yourself:

int contiguous_matrix[81];

int& location(int row, int column)
{
  return contiguous_matrix[row * 9 + column];
}

You can also iterate over each column of a row:

typedef void (*Function_Pointer)(int&);

void Column_Iteration(Function_Pointer p_func, int row)
{
  row = row * MAXIMUM_COLUMNS;
  for (unsigned int column = 0; column < 9; ++column)
  {
    p_func(contiguous_matrix[row + column]);
  }
}

Answer 7

For static declared arrays you can access them like continuous 1D array, p = matrix[0] will give you the 1st column of 1st row. Then the 1D array can be accessed like p[i] , *(p+i) , or p[current_raw * raw_size + current_column) .

The things are getting tricky if a 2D array is represented with **p as it will be interpreted as an array of pointers to 1D arrays.

Answer 8

I don't know if this is a efficient solution but by doing this way I'm able to get the column,

int arr[9][2] = { {2,  57}, {3,  66}, {4,  73}, {5,  76}, {6,  79}, {7,  81}, {8,  90}, {9,  96}, {10, 100}};

int c[18];
int co = 0;
for (auto & i : arr) {
    for (int j : i) {
        c[co++] = j;
    }
}

for (int i = 0; i < co; ++i) {
    if (i % 2 != 0)
        std::cout << c[i] << " ";
}

Answer 9

#include <iostream>

using namespace std;

int main()
{
   int aa[10][10];

   for(int i = 0; i<10; i++)
       for(int j = 0; j<10; j++)
           aa[i][j] = i*10+j;
   
   int col = 2;

   // pointer to a length-10 1d array
   int (*p)[10] = (int (*)[10])&(aa[0][col]);

   for(int i =0; i<10; i++)
      cout << *(p[i]) << endl;

   return 0;
}

ARRAY ADDRESS AND POINTERS TO MULTIDIMENSIONAL ARRAYS