starting address of array a and &a

Question

In the below two lines,

char a[5]={1,2,3,4,5};
char *ptr=(char *)(&a+1);
printf(\"%d\",*(ptr-1));

This prints 5 on screen.Whereas when use

Answer 1

What you are running into is a subtlety of pointer arithmetic.

The compiler treats "a" as a pointer to char - an entity that is 1 byte in size. Adding 1 to this yields a pointer that is incremented by the size of the entity (i.e. 1).

The compiler treats "&a" as a pointer to an array of chars - an entity that is 5 bytes in size. Adding 1 to this yields a pointer that is incremented by the size of the entity (i.e. 5).

This is how pointer arithmetic works. Adding one to a pointer increments it by the size of the type that it is a pointer to.

The funny thing, of course, is that when it comes to evaluating the value of "a" or "&a", when dereferencing, they both evaluate to the same address. Which is why you see the values that you do.