Memory allocation for pointer to array and array of pointer

Question

Below is the part of code I am stuck on. I want to dynamically allocate memory for

• Pointer to array
• Array of pointers

I am getting several error messages like invalid conversion from int * to int and so on.

Pointer to array

int (*array)[nrows][ncolumns];
array = (int*)malloc(nrows * ncolumns * sizeof(int));
printf("\n Enter the elements:\n");
for(i=0; i<nrows; i++)
{
 for(j=0; j<ncolumns; j++)
 {
 scanf("%d", array[i][j]); 
 }
}
printf("Entered array is :\n\n");
for(i = 0;i<nrows; i++)
{
 for(j = 0; j<ncolumns; j++)
 {
 if(j== ncolumns-1)
 {
 printf("%d \n", *array[i][j]);
 }
 else
 {
 printf("%d", *array[i][j]);
 }

Array of pointers

int *array[nrows][ncolumns];
array[nrows][ncolumns] = (int*)malloc(nrows * ncolumns * sizeof(int));
printf("Enter elements:\n");
for(i = 0; i<nrows; i++)
{
 for(j = 0; j<ncolumns;j++)
 {
 scanf("%d",&array[i][j]);
 }
}
printf("Entered array is: \n");
for(i = 0; i<nrows; i++)
{
 for(j = 0; j<ncolumns;j++)
 {
 if(j == ncolumns-1)
 {
 printf("%d \n",array[i][j]);
 }
 else
 {
 printf("%d \t",array[i][j]);
 }
 }
}

• molbdnilo: I recommend that you start with one-dimensional arrays until you're comfortable with the difference between arrays and pointers, and the difference beween a pointer to an array and a pointer to an array's first element. Also, that you stay away from mallocing these things until you can write the code without it.
• SKD: Thanks. But actually I don't have that much time. I have to correct it in 1 hour. Please help me out. Is the memory allocation part correct?
• molbdnilo: @Sujit Well, you should have began studying earlier. Hints: In 1), array[i] is a two-dimensional array, array[i][j] is a one-dimensional array. In 2), array[i][j] is an (uninitialised) pointer, not an int.
• molbdnilo: @Sujit array[nrows][ncolumns] is an element outside the array; assigning to it is undefined.
• M.M: @Sujit you also need to fix the bugs in your use of scanf and printf, it should be scanf("%d", &array[i][j]); and printf("%d", array[i][j]);

Answer 1

1> pointer to array

#include <stdio.h>
#include <stdlib.h>
int main(void){
 int nrows = 3;
 int ncolumns = 4;
 int i, j;
 int (*array)[nrows][ncolumns];//do you want <<int (*array)[ncolumns]>> ?
 //like as int src[nrows][ncolumns]; array = &src;
 array = malloc(nrows * ncolumns * sizeof(int));//(int*) : type mismatch
 printf("\nEnter the elements:\n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns; j++){
 scanf("%d", &(*array)[i][j]);
 }
 }
 printf("Entered array is :\n\n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns; j++){
 if(j != 0)
 putchar(' ');
 printf("%d", (*array)[i][j]);//need ( )
 }
 putchar('\n');
 }
 free(array);
 return 0;
}

---

2> Array of pointers

#include <stdio.h>
#include <stdlib.h>
int main(void){
 int nrows = 3;
 int ncolumns = 4;
 int i, j;
 int *array[nrows][ncolumns];
 int *src = (int*)malloc(nrows * ncolumns * sizeof(int));//no need (int*)
 printf("Enter elements:\n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns;j++){
 array[i][j] = &src[ i * ncolumns + j];//pointer pointed to entity (src[ i * ncolumns + j])
 scanf("%d", array[i][j]);//type of array[i][j] is int *
 }
 }
 printf("Entered array is: \n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns; j++){
 if(j != 0)
 putchar(' ');
 printf("%d", *array[i][j]);//need * for dereference
 }
 putchar('\n');
 }
 free(src);
 return 0;
}

---

3> option

#include <stdio.h>
#include <stdlib.h>
int main(void){
 int nrows = 3;
 int ncolumns = 4;
 int i, j;
 int (*array)[ncolumns];
 array = (int (*)[ncolumns])malloc(nrows * sizeof(*array));//sizeof(*array) : sizeof(int[ncolumns])
 printf("\nEnter the elements:\n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns; j++){
 scanf("%d", &array[i][j]);
 }
 }
 printf("Entered array is :\n\n");
 for(i = 0; i<nrows; i++){
 for(j = 0; j<ncolumns; j++){
 if(j != 0)
 putchar(' ');
 printf("%d", array[i][j]);
 }
 putchar('\n');
 }
 free(array);
 return 0;
}

• SKD: Thanks a lot. It really helped. Though, in 1st program, M still getting the 'Type conversion' error..
• BLUEPIXY: @Sujit Perhaps you are using C++ compiler as C compiler. (cast from void * no need in C ), try array = (int (*)[nrows][ncolumns])malloc(nrows * ncolumns * sizeof(int));
• SKD: I am using Dev-C++ compiler. Thanks once again. Now both progrmas are running well.
• SKD: Means? Where to add & which option?
• BLUEPIXY: @Sujit add example of 3> option. M.M's comment for int (*array)[ncolumns]