This was an exam problem that I still don't really understand
Asking for the outputCode:int main () int a[10] = {1,2,3,4,5,6,7,8,9,10}; int *ptr = (int*)(&a + 1); printf("%d %d, *(a+1), *(ptr-5) ); return 0;
I understand that *(a+1) == 2
but
how does *(ptr-5) == 6?
It makes sense to me that *(ptr+5) == 6.. but dont know how *(ptr-5) == 6?
thanks in advance
Try print the result of this line
printf("%p %p\n", &a+1, a+1);
The type of &a is not the same as a.
&a+1 is just passed the end of the array a
a+1 is the second element in the array a
Tim S.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
Right, &a is pointer to int[10]. incrementing a pointer moves it by size of the type it points to, in this case sizeof(a) is 10*sizeof(int). The rest is strait forward.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
