int * p = (void *) 1 seg fault

int * p = (void *) 1; /* seg fault. why is this happening ? */

Do you actually expect "1" to be a valid memory address?

The first MiByte of RAM in modern PCs is always reserved by the system for special purposes (old device support, for example).

/*more specific*/

Code:

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int * ptr = (void*) 1;
printf("The value of ptr is %i\n",ptr);  /*why this is legal*/
printf("The value of ptr is %ls\n",ptr); /*why this is seg fault*/

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Are you trying to learn C just by trying stuff to see what crashes and what doesn't?
The problem with this approach is there is no guarantee of failure when you do something wrong.

> printf("The value of ptr is %ls\n",ptr); /*why this is seg fault*/
Not only is it dumb to try and dereference an int pointer to random memory locations, it's even dumber still to try and print it as a string.

This about summarises what you can do with an int pointer in a typical user-space program running on any common OS.

Code:

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#include <stdio.h>
#include <stdlib.h>
int main ( ) {
    struct foo {
        int bar;
    } structvar;
    int arrayvar[10];
    int intvar;
    int *ptr;

    ptr = NULL;            // valid, but you can't dereference it
    ptr = &intvar;          // *ptr is valid

    ptr = arrayvar;        // *ptr is valid, as is ptr[0] tru ptr[9]
    ptr = &arrayvar[0];    // *ptr is valid, as is ptr[0] tru ptr[9]
    ptr = &arrayvar[5];    // *ptr is valid, as is ptr[0] tru ptr[4]
    ptr = &arrayvar[10];    // valid, but you can't dereference it.
                            // You can however use <, != comparisons with another
                            // ptr to the same array

    ptr = &structvar.bar;  // *ptr is valid

    ptr = malloc( sizeof(*ptr));    // *ptr is valid
    free(ptr);              // Only free what malloc returned.
}

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