Question

I was revisiting pointers when I had this doubt.

int *ptr;
int arr[5] = {10,20,30,40,50};
ptr = &arr[0];

Now printf("Value: %d",*ptr); would print 10

if I do:

ptr++;
printf("Value: %d",*ptr);

the result would be: 20

Similarly if I do:

*ptr++;
printf("Value: %d",*ptr);

the result is: 30

But

printf("Value: %d",++*ptr);

gives me 31

I thought since ptr is of type int when I increment it, it would jump 4 bytes to the next memory location. But why does it show the same behavior for *ptr++ and ptr++ and not for ++*ptr?

Answer 1

The first snippet is obvious: it prints what ptr points to, i.e. 10.

The second one, moves the pointer forward of one element, which then points to the next element, i.e. 20.

The third snippet is exactly the same as the previous one, because its first instruction increments the pointer and returns the unincremented value, which is dereferenced, but its result is discarded; what is dereferenced in the printf is the incremented pointer, which now points to 30.

The last snippet is different: ++*ptr is ++(*ptr); *ptr dereferences ptr (which already points to 30), yielding 30, and ++ increments such value, which becomes 31.