Dynamic 2D Array in C++ with explanation

Dynamic 2D Array In C++


Try to create integer 2D array of 4 row and 3 columns.

Code snippet:
int** a = new int* [4];//line1
for(int i = 0; i < 4; i++) {//line2
a[i] = new int[3];//line3
} //line4


Explanation:
Let, Address of a is 0x100 and size of int is 4.
So, Address of a[0] is 0x100, address of a[1] is 0x104, address of a[2] is 0x108 and so on.
a[0]                             a[1]                        a[2]                          a[3]



0x100                       0x104                      0x108                     0x10C

1st Iteration of line3. Assume receives starting address is 0x500. So, three address is 0x500,0x504,0x508.
a[0] = 0x500;



0x500                           0x504                           0x508 
Updated value of a is:
0x500



0x100                   0x104                   0x108                 0x10C
2nd Iteration of line3. Assume receives starting address is 0x600. So, three address is 0x600,0x604,0x608.
a[1] = 0x500;



0x600                            0x604                             0x608              
Updated value of a is:
0x500
0x600


0x100                   0x104                   0x108                     0x10C

3rd Iteration of line3. Assume receives starting address is 0x700. So, three address is 0x700,0x704,0x708.
a[2] = 0x700;



0x700                              0x704                               0x708         
Updated value of a is:
0x500
0x600
0x700

0x100                   0x104                     0x108                    0x10C

4th Iteration of line3. Assume receives starting address is 0x800. So, three address is 0x800,0x804,0x808.
a[3] = 0x800;



0x800                               0x804                                0x808        
Updated value of a is:
0x500
0x600
0x700
0x800
0x100                     0x104                      0x108                     0x10C






Try to put value in array from 1 to 12.
Code snippet
int x = 0; //line5
for (int i = 0; i < 4; i++) {//line6
for (int j = 0; j < 3; j++) {//line7
       a[i][j] = ++x;//line8
   }
}

a[i][j] -> *(*(a+i)+j)

For 1st iteration of i:
     For 1st iteration of j:
a[0][0] -> *(*(a+0)+0)
a= 0x100;
a+0 ->0x100;
*(a+0) -> 0x500
(*(a+0)+0) ->0x500
*(*(a+0)+0) ->1
Then, a[0][0] = 1;
    
For 2nd iteration of j:
a= 0x100;
a+0 ->0x100;
*(a+0) -> 0x500
(*(a+0)+1) ->0x500+1*sizeof(int)=0x500+4=0x504
*(*(a+0)+1) = 2
Then, a[0][1] = 2;

Going so on.
Let, 4th iteration of i and 3rd iteration of j
a[3][2] = 12;
a[3][2] -> *(*(a+3)+2)
a=0x100;
a+3 -> 0x100 + 3*sizeof(int) = 0x100 + 3*4 = 0x10C
//Hex value 0x10C = Decimal value 112.
*(a+3) = *(0x10C) -> 0x800
(*(a+3)+2) = 0x800 + 2*sizeof(int) = 0x800+2*4=0x808
*(*(a+3)+2) = 12 // value 0x808 address

1
2
3
0x500                               0x504                                0x508      
  
4
5
6
0x600                              0x604                                 0x608        

7
8
9
0x700                                  0x704                              0x708       

10
11
12
0x800                               0x804                                  0x808      

  

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