Question

How does a pointer points to [-1]th index of the array produce legal output everytime. What is actually happening in the pointer assignment?

#include<stdio.h>
int main()
{
        int realarray[10];
        int *array = &realarray[-1];

        printf("%p\n", (void *)array);
        return 0;
}

Code output:

manav@workstation:~/knr$ gcc -Wall -pedantic ptr.c
manav@workstation:~/knr$ ./a.out
0xbf841140

EDIT: If this scenario is valid, then can i use this to define an array whose index start from 1 instead of 0, namely: array[1], array[2],…

Answer 1

Youre simply getting a pointer that contains the address of that “imaginary” location, i.e. the location of the first element &realarray[0] minus the size of one element.

This is undefined behavior, and might break horribly if, for instance, your machine has a segmented memory architecture. It’s working because the compiler writer has chosen to implement the arithmetic as outlined above; that could change at any moment, and another compiler might behave totally differently.