Significance of double pointer and triple pointer in this code snippet - c

Question

#include<stdio.h>
#include<stdlib.h>
void add(char **p);
void print(char **p);
int cnt=0;
main()
{
int option;
char **p=NULL;
while(1)
{
printf("------MENU-----\n");
printf("1>input\n 2>print\n3>exit\n");
printf("enter ur choice\n");
scanf("%d",&option);getchar();
switch(option)
{
case 1: add(p);
break;
case 2: print(p);
break;
case 3: return;
default: printf("Invalid option\n");
}
}
}
void add(char **p)
{
int i;
p=(char**)realloc(p,(cnt+1)*sizeof(char*));
if(p==NULL)
{
printf("Error: memory not available\n");
return;
}
p[cnt]=NULL;
p[cnt]=(char*)realloc(p[cnt],20*sizeof(char));
puts("enter a name");
gets(p[cnt]);
cnt++;
printf("cnt=%d\n",cnt);
}
void print(char **p)
{
int i;
for(i=0;i<cnt;i++)
printf("p[%d]=%s\n",i,p[i]);
}
In the above code, I am making a database of names. For this I am using dynamic memory allocation. I am allocation memory for 2D-array using array of pointers method. When I am executing this program on gcc compiler, I am getting segmentation fault. I am not understanding why is it happening?Could you please tell me where the bug is?

Answer 1

In main, all you do is to assign p = NULL and then use this NULL pointer in print, which causes the segfault. Note that add(p) is equivalent to add(NULL) and does not change p.
Did you mean to pass an address with add(&p)? If so, you need to fiddle a bit with the number of *.

Answer 2

p in main is handed over to add by value. add modifies a local copy then, but not the original p.
Besides the terrible formatting and everything you need to hand a pointer to your main's p to add:
...
case 1: add(&p);
...
void add(char ***p)
{
int i;
*p = realloc(*p,(cnt+1)*sizeof(char*));
if(*p==NULL)
{
printf("Error: memory not available\n");
return;
}
(*p)[cnt]=NULL;
(*p)[cnt]=realloc((*p)[cnt],20*sizeof(char));
puts("enter a name");
gets((*p)[cnt]);
cnt++;
printf("cnt=%d\n",cnt);
}

Answer 3

A copy of 2D pointer p is created in the add function, and updating this pointer will not be reflected in the main function.
A quick solution to this problem is to return the 2D pointer 'p' from the add function.
char** print(char **p)
{
....
....
return p;
}

Answer 4

Pass by reference. You need to pass the address of your pointer to make sure the changes in function add() is reflected in main(). Avoid using gets()
In this case
add(&p);
Accordingly your add() function definition should change to handle this.
Or the other way is for the add function to make the required allocations and return that address to your pointer
char **add();
Check the code below:
char **add(char **p)
{
int i;
p=(char**)realloc(p,(cnt+1)*sizeof(char*));
if(p==NULL)
{
printf("Error: memory not available\n");
return;
}
p[cnt]=NULL;
p[cnt]=(char*)realloc(p[cnt],20*sizeof(char));
scanf("%s",p[cnt]);
cnt++;
printf("cnt=%d\n",cnt);
return p;
}
So your call should be:
p = add(p);

Answer 5

Apart from the other things, I don't think anyone's mentioned the realloc trap of reallocation failure. Suppose this call
p=(char**)realloc(p,(cnt+1)*sizeof(char*));
fails to allocate the new memory: what happens then? Yes, you will get NULL returned: but the memory which was allocated, and which pis pointing to, does not get free'd. Instant memory leak.
You must call realloc like this:
char* pTemp = realloc(p, cnt + 1);
if(pTemp != NULL)
{
p = pTemp; // Now safe to update p
}
else
{
// Error handling as required, possibly including freeing p
}
Yet another reason not to use realloc. It doesn't really by you very much over doing the buffer copies yourself IMHO.