Question

#include<stdio.h>
int main(){

char a[6],*p;

a[0]='a';
a[1]='b';
a[2]='c';
a[3]='4';
a[4]='e';
a[5]='p';
a[6]='f';
a[7]='e';
printf("%s\n",a);
printf("printing address of each array element");
p=a;

printf("%u\n",&p[0]);
printf("%u\n",p+1);
printf("%u\n",a+2);
return 0;
}

The output is as follows…

anusha@anusha-laptop:~/Desktop/prep$ ./a.out
abc4epfe
printing address of each array element3216565606
3216565607
3216565608

When I declared the array as char a[6] why is it allowing me to allocate a value at a[7]? Does it not need a null character to be appended for the last element?

Also p=a => p holds the address of first element of char array a. I don’t understand how it is correct to place an ‘&’ in front of an address (p[0]). &p[0] means address of address of first element of a which doesn’t make any sense, at least to me.

Why is it printing the correct output?

Answer 1

You have just invoked undefined behaviour. There’s little point in reasoning about writing beyond the bounds of an array. Just don’t do it.

&p[0] means address of address of first element of a[] which is not sensible

No, that’s perfectly sensible. Your description perfectly describes what is going on. &p[0] is the same as p which is the same as a. When you write p[0] you are dereferencing the pointer. When you then write &p[0] you are taking the address of that variable and thus return to what you started from, p.