Memory assignment in C through pointers

I am learning how to use pointers, so i wrote the below program to assign integer values in the interval [1,100] to some random locations in the memory.

When i read those memory locations, printf displays all the values and then gives me a segmentation fault. This seems an 开发者_如何转开发odd behavior, because i was hoping to see either all the values OR a seg fault, but not both at the same time.

Can someone please explain why i got to see both?

Thanks. Here is the code

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

int main()
{
    char first = 'f';
    char *ptr_first = &first;
    int i=1;
    for(i=1;i<101;i++)
        *(ptr_first+i) = i;
    for(i=1;i<101;i++)
        printf("%d\n", *(ptr_first+i));
    return EXIT_SUCCESS;
} 

Not odd at all. You are using your variable first, which is on the stack. What you essentially do is happily overwriting the stack (otherwise known from buffer overflows on the stack) and thus probably destroying any return address and so on.

Since main is called by the libc, the return to libc would cause the crash.

You're accessing memory past beyond that assigned to first. It is just one character, and, through the ptr_first pointer, you're accessing 100 positions past this character to unreserved memory. This may lead to segfaults.

You have to ensure the original variable has enough memory reserved for the pointer accesses. For example:

char first[100];

This will convert first in an array of 100 chars (basically a memory space of 100 bytes that you can access via pointer).

Note also that you're inserting int into the char pointer. This will work, but the value of the int will be truncated. You should be using char as the type of i.

since ptr_first pointer is pointing to a char variable first. Now when you are incrementing ptr_first, so incremented memory address location can be out of process memory address space, thats why kernel is sending segmentation fault to this process.