C Pointer casting: single int pointer to double char pointer

Question

I am trying to do the following, it does not give me compilation error but gives me a segmentation fault when I run this.

int main(){
int * ptr;
*ptr = 254;
char **ch = (char **)ptr;
printf("\n%x\n",**ch);
}

I basically want to know if something like this is legal, and if so what it does.

*(char **)ptr

where ptr is either of type int or type void

Answer 1

*ptr = 254; here ptr is not allocated memory, that's why producing segmentation fault.

First you need to allocate the memory to ptr , then you can put some value into *ptr .

Next, *(char **)ptr is not legal (neither is char **ch = (char **)ptr; , BTW). If you compile with warnings enabled, you'll get some warning message

warning: initialization from incompatible pointer type

You can easlity understand this if you try to analyze the data type of either variables. Consider a sample code

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

    int main(){
            int * ptr = malloc(4);
            *ptr = 254;
            char **ch = (char **)ptr;
            printf("\n%x\n",**ch);
    }

To check, if you compile and step through a debugger, you can see,

5           int * ptr = malloc(4);
(gdb) s
6           *ptr = 254;
(gdb) 
7           char **ch = (char **)ptr;
(gdb) p ptr
$1 = (int *) 0x804a008
(gdb) p ch
$2 = (char **) 0xa7c600
(gdb) p *ch
$3 = 0x57e58955 <Address 0x57e58955 out of bounds>
(gdb) p **ch
Cannot access memory at address 0x57e58955

Answer 2

The variable ptr is uninitialized.

A pointer is a variable which stores address.

Here the variable ptr is not initialized to any value.

When it tries to access ptr it crashes for sure.

Because ptr has a junk address value, and you are trying to access that.

With the given code, that must be the issue.

But the issue is not with the type conversion to char**.

Answer 3

Few things needs to be set right in your code:

1. ptr is a pointer and it should point to some memory location in your case it is not doing so.So you need to allocate some memory and then you need to access the pointer.
2. You are trying to use a character pointer to point to integer pointer. Now this character pointer will have only 1 byte of data to access and not more than that.(Size of integer variable may vary on different platform).

Take a look at the below code: Let's say you assign some memory dynamically using malloc or use a pointer to point to some location as shown below.

#include<stdio.h>
#include<string.h>
   int main(){
      int a =10;
      int * ptr = &a; 
      *ptr = 81;/* ASCII value of Q is 81 */
      printf("%d\n",a);
      char **ch = (char **)&ptr;/* Pointer pointing to a pointer */
      printf("%c\n",**ch);/* You will see Q getting printed */
      /* Now if you want to modify the value of a it can be done */
      **ch = 'R';
      printf("%d\n",a); /* The value stored in a will be 82 ASCII value of R */
      printf("%c\n",**ch);
      printf("%d\n",**ch); /* You should get 82 here */
      /* Now lets say you try to set a to value 290 */
      a = 290;
      printf("%d\n",a);/* You get 290 here */
      printf("%d\n",**ch); /* Sorry you won't get 290 here */ 
      printf("%c\n",**ch); 
      /* Since only 1 byte is what char pointer can access you get equivalent ASCII value for the value matching 8 bits */
   }