1

Can we dynamically allocate 2D array without using any for loop or while loops? i there any direct command or function in c c++?

Aragon
  • 1,551
  • 1
  • 15
  • 25

6 Answers6

4

Without using a loop you will have one restriction in ISO c++ i.e. size of one dimension has to be determined at compile time. Then this allocation can be done in a single statement as follows:

#define COLUMN_SIZE 10 // this has to be determined at compile time
int main()
{
    int (* arr)[COLUMN_SIZE];
    int rows = 20; // this is dynamic and can be input from user at run time
    arr = new int[rows][COLUMN_SIZE];
    arr[3][4] = 10;
    cout << arr[3][4] << endl;
    return 0;
}

The memory allocated with new needs to be freed. Also if we extend it to n dimensions, only one of these dimensions can be determined at run time. The reason is that compiler has to know the size of each row in order to create a row of contiguous memory.

fkl
  • 5,412
  • 4
  • 28
  • 68
1

Although you should avoid raw pointers, this should work->

int *myArray = new int[R*C];

Here R is number of rows and C is number of columns. Although it is really a 1D array, you can manipulate it as 2D array. For example, myArray[i][j] can be read as->

myArray[i*C + j]
Vinayak Garg
  • 6,518
  • 10
  • 53
  • 80
  • Yeah but effective syntax is still of a 1D array and it cannot be accessed using [][] or *(*(arr+row)+col) – fkl Sep 28 '12 at 07:30
  • 2
    And in fact this is almost always what you want when talking about a 2D array in constrast to an array of arrays (which is probably what the OP confused with a 2D array). – Christian Rau Sep 28 '12 at 07:34
  • If you want to avoid raw pointers, then `vector a(R*C)` obviously has the same effect. – Fred Foo Sep 28 '12 at 08:26
0

The only way to do it with out loops is to allocate a psuedo 2D array thus:

int *ary = new int[sizeX * sizeY];

but then accessing that is non standard & frankly ugly:

ary[y*sizeX + x]

If you want a "real" 2D array then your stuck with loop intialization:

int **ary = new int*[sizeY];
for(int i = 0; i < sizeY; ++i) {
    ary[i] = new int[sizeX];
}

But then you have to be careful about clean up:

for(int i = 0; i < sizeY; ++i) {
    delete [] ary[i];
}
delete [] ary;

So in my view

std::vector<std::vector < int> >

is probably the simplest and safest way to go in a real world app.

Ricibob
  • 7,505
  • 5
  • 46
  • 65
  • Pseudo 2D array is not the only way. You can still allocate without loop provided one dimensions value is determined at compile time. – fkl Sep 28 '12 at 07:32
0

Alternative way to access in arr[..][..] format.

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

int main()
{
    int COL ;
    int ROW ;
    COL = 8;
    ROW = 12;
    int (*p)[COL];
    int *mem = (int*)malloc(sizeof(int)*COL*ROW);
    memset(mem,0,sizeof(int)*COL*ROW);
    p = (int (*)[10])mem;

    printf("0x%p\n", p);
    printf("0x%p %d\n", p+1, (((int)(p+1))-((int)p))/sizeof(int));

    mem[2*COL+0] = 1;
    printf("%d\n", p[2][0]);
    mem[2*COL+5] = 2;
    printf("%d\n", p[2][5]);
    mem[6*COL+7] = 3;
    printf("%d\n", p[6][7]);

    p[1][2] = 4;
    printf("%d\n", mem[1*COL+2]);

    free(p);

    return 0;
}

Of course, you can do int (*p)[COL] = (int (*)[COL]) malloc(sizeof(int)*COL*ROW); directly.

Marcus
  • 6,701
  • 4
  • 19
  • 28
-1

Try to replace loop by recursion

Aleksey Bakin
  • 1,506
  • 13
  • 27
-1

A std::map<TypeDim1, std::map<TypeDim2, TypeContent> > might be a dynamically allocated choice to represent a 2D array.

#include <map>
typedef std::map<int, std::map<int, std::string> > array2dstring;

int main(int argc, char *argv[])
{
   array2dstring l_myarray2d;
   l_myarray2d[10][20] = "Anything"; 
}
Uli Klank
  • 199
  • 10