array of pointers

Question

char *a[]={"diamonds","clubs","spades","hearts"};
char **p[]={a+3,a+2,a+1,a};
char ***ptr=p;
cout<<*ptr[2][2];

why does it display h and please explain how is the 2d array of ptr implementing and its elements

Answer 1

Note that x[y] binds tighter than *x, so the expression *ptr[2][2] is interpreted as *(ptr[2][2]).

Also note that x[y] == *(x+y).

Therefore

*(ptr[2][2]) == *(p[2][2])    // ptr = p
             == *((a+1)[2])   // p[2] == a+1
             == *(*(a+1+2))   // x[y] == *(x+y)
             == *(*(a+3))     // 1+2 == 3
             == *(a[3])       // *(x+y) == x[y]
             == *("hearts")   // a[3] == "hearts"
             == "hearts"[0]   // *x == *(x+0) == x[0]
             == 'h'

Answer 2

See KennyTM's excellent answer for an explanation... but I thought this would be the perfect case for professing the use of your debugger to "visualize" memory.. and provide an easy answer to this type of question.