Point out the correct statement will let you access the elements of the array using 'p' in the following program?

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

int main()
{
    int i, j;
    int(*p)[3];
    p = (int(*)[3])malloc(3*sizeof(*p));
    return 0;
}

    for(i=0; i<3; i++)
    {
       for(j=0; j<3; j++)
       printf("%d", p[i+j]);
    } 
    

    for(i=0; i<3; i++
         printf("%d", p[i]);
    

    for(i=0; i<3; i++)
    {
       for(j=0; j<3; j++)
            printf("%d", p[i][j]);
    }        
    

    for(j=0; j<3; j++)
      printf("%d", p[i][j]);      
    

Point out the error in the following program.

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

int main()
{
    char *ptr;
    *ptr = (char)malloc(30);
    strcpy(ptr, "RAM");
    printf("%s", ptr);
    free(ptr);
    return 0;
}

Point out the error in the following program.

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

int main()
{
    int *a[3];
    a = (int*) malloc(sizeof(int)*3);
    free(a);
    return 0;
}

How many bytes of memory will the following code reserve?

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

int main()
{
    int *p;
    p = (int *)malloc(256 * 256);
    if(p == NULL)
        printf("Allocation failed");
    return 0;
}

Assume integer is 2 bytes wide. What will be the output of the following code?

#include<stdio.h>
#include<stdlib.h>
#define MAXROW 3
#define MAXCOL 4

int main()
{
    int (*p)[MAXCOL];
    p = (int (*) [MAXCOL])malloc(MAXROW *sizeof(*p));
    printf("%d, %d\n", sizeof(p), sizeof(*p));
    return 0;
}