Making a two dimensional array at run time using malloc

Question

I have been trying to allocate a 2 dimensional array at run time using malloc by using the concept of pointers to pointers. This code doesn't show any compilation error but gives a runtime error.

#include<stdio.h>
#include<stdlib.h>
int ** alpha(void)
{
 int **x;
 x=(int **)malloc(3*sizeof(int *));
 for(int i=0;i<3;i++)
 {
 x[0]=(int *)malloc(3*sizeof(int));
 }
 for(int i=0;i<3;i++)
 {
 for(int j=0;j<3;j++)
 {
 x[i][j]=i+j;
 }
 }
 return x;
}
int main()
{
 int **p;
 p=alpha();
 printf("%d",p[1][2]);
 return 0;
}

• Alex Celeste: There's a difference in C between a two-dimensional array and an array-of-arrays, btw. This is the latter.
• AndersNS: And the run time error is..?

Answer 1

The reason for the runtime error is because you allocate memory just for x[0].So, change

x[0]=(int *)malloc(3*sizeof(int));

to

x[i]=malloc(3*sizeof(int));

and don't cast the result of malloc.

Answer 2

In

x[0]=(int *)malloc(3*sizeof(int));

you ended up allocating memory for only index 0 three times. use i.

After you're done using , don't forget to free() the allocated memory. Otherwise, it'll result in a memory leak.

Also, no need to cast the return value of malloc()/calloc().