/*..AN IMPLEMENTATION TO REPRESENT 3D ARRAY..

/*..AN IMPLEMENTATION TO REPRESENT 3D ARRAY..*/

//HERE I IS THE ARRAY IN WHICH POSITIONS  ACCORDING TO J ARE LEFT...THRN J IS AGAIN  AN ARRAY IN WHICH POSITIONS ACCORDING TO K ARE LEFT....

//MEANS ULTIMATELY I WILL LEFT POSITIONS ACCORDING TO K

//SUPPOSE I IS ++MENTING IN CONTINOUS FORM AFTER LEAVING 40 SPACES ...

//means arr[10][10][10] is asking for k before to leave space of 4-4 bytes 10 times to put 10 values..in array ...so diffrence between addresses when i is changing is 40 bytes..//

#include<stdio.h>

int main()

{

int i,j,k,arr[10][10][10];

int x,y,z;

printf("PLEASE ENTER VALUE LINES IN X-AXIS Y-AXIS & Z-AXIS:\n");

scanf("%d%d%d",&x,&y,&z);//[0][0][0]..[0][0][1]...[0][0][2]...  //[0][1][0]..[0][1][1]..[0][1][2]..   .//[1][0][0]..[1][0][1]..[1][0][2].... //[1][1][0]...[1][1][1]...[1][1][2]..//

for(i=0;i<x;i++)//FOR EACH LOOP OF i...//

{

for(j=0;j<y;j++)//FOR EACH VALUE OF I ...J ++MEANTS

{

for(k=0;k<z;k++)//FOR EACH VALUE OF J..K ++MENTS..//

{

printf("[%d][%d][%d]=",i,j,k);

scanf("%d",&arr[i][j][k]);//ENTER VALUE FOR EACH OF ABOVE OPERATING LOOPS...MEANS FOR 000..001..002..010..011...012..100..101...102...110...111...112..//

}

printf("\n");//PRINT INTO NEW LINE

}printf("\n");//print into new line//

}

for(i=0;i<x;i++)

{

for(j=0;j<y;j++)

{

for(k=0;k<z;k++)

{

printf("[%d][%d][%d]=",i,j,k);

//printf("%d",*(arr+i));//will print the base address of array...each time in loop//

//printf("%d",**(arr+i+j));//will also print the base address of array...each time in loop//

//printf("%d",*(*(arr+i)+j));//will print the base address of each newarray...each time in loop//

printf("%d",*(*(arr+i)+j)+k);//will print the address of each new element in array...each time in loop//

//printf("%d",*(*(*(arr+i)+j)+k));//print the value in array

}

printf("\n");

}

printf("\n");

}

return 0;

}