confusing on pointer and array

Question

We have

 int a[5]={10, 20, 30, 40, 50};

I would like to know how does the following two code segment do?

 int *ptr = (int *)(&a+1);
 int *t = (int *)(&a -1);

If we have

 printf("%d  %d  %d \n", *(a+1), *(ptr-1), *(t+1));

What should be the result?

Answer 1

Since the type of a is array-of-5-ints, that means that the type of &a is pointer-to-array-of-5-ints.

When you add or subtract 1 from a pointer, you ask it to point to the next or previous object of that type in memory. So &a+1 is creating a pointer to the array-of-5-int immediately after a in memory (which doesn't exist), and &a-1 is creating a pointer to the array-of-5-int immediately before a in memory (which also doesn't exist). In memory, it looks like this (where each cell represents one int):

Address:    &a-1                      &a                      &a+1
Contents:  | ?  | ?  | ?  | ?  | ?  | 10 | 20 | 30 | 40 | 50 | ?  | ?  | ?  | ?  | ?  |

When a is used in the expression *(a+1), it is converted to a pointer to its first element - so a pointer-to-int pointing at the 10 value. Adding one to it then makes a pointer pointing at the next int - a+1 points at the 20 value. *(a+1) then fetches that value, so the first number printed is 20.

As ptr is also a pointer-to-int, that means that ptr - 1 creates a pointer to the int immediately before ptr - in this case, it'll be pointing at the 50. So the second number printed is 50.

Similarly, t + 1 creates a pointer to the int immediately after t - in this case, it's the second ? in the above diagram. This is an uninitialised value - it could print anything at all, or even crash the program.

Address:    &a-1                      &a                       &a+1
            t    t+1                  a   a+1            ptr-1 ptr
Contents:  | ?  | ?  | ?  | ?  | ?  | 10 | 20 | 30 | 40 | 50  | ?  | ?  | ?  | ?  | ?  |