Programming

Thanks Shamrock!
Good hints are So, a[1] is:{{'g', 'h'}, {'i', 'j'}, {'k', 'l'}}
......
Read more last night, and finally I got the answers, just for somebody who hits the same wall as I did.

as (*pa)[2] = &a[1][0], which means *pa is a "2-col" pointer start from {'g', 'h'} in the array a; so that char 'r' is in {'q', 'r'}, the 6th member of pa ---> pa[5], which in turn is a "2-char" array, and with the second index = 1. So the answer is pa[5][1];
as (*ppa)[3][2] = &a[1], which defines *ppa is a "3(row) x 2(col) array " pointer start from {{'g', 'h'}, {'i', 'j'}, {'k', 'l'}} in the array a. For char 'r' is on the 2nd row, 3rd col, within the 3rd col it is in the 2nd col. So the answer is ppa[1][2][1].

The reason why there is multiple answer to get the 'r' is because C actually only has "1-dimension" or linear arrangement of the array members in memory as Corona688 pointed out. When you split the linear array in different combinations, counted by 1, by 2, or by 3, 4, 5.... Actually pa[11] prints 'r', pa[3][2], pa[4][3] pa[5][1]...all point to 'r', but only pa[5][1] is appropriate to the original question. Similar for ppa, but only pa[1][2][1] is the correct answer as it was defined as a 3x2 array.

Shamrock and Corona688, please correct me if I interpreted anything wrong.
Thanks a lot again!

---------- Post updated 09-25-13 at 12:50 PM ---------- Previous update was 09-24-13 at 05:50 PM ----------

To confirm my digestion I tried another two situations by change the pa and ppa definitions:
the answers to get char 'r' are

Code:

pa[7][1]; 
ppa[-1][2][1]

Here it is very interesting the index can be negative. Amazing C!

That's just how the processor, and two's-complement math works, not a special property of C per-se. It lets you do so, when most languages would protect you from it, but C, which hardcodes so much into assembly language at compile-time (same reason sizeof() can't be used dynamically) is very limited in what it can warn you about.

In most situations, a negative array index would edge into incorrect or even invalid memory, causing strange behavior or crashes. It's only okay here since you know you're not at the beginning of the data of interest.

So, use with care.

Sure! It was only to confirm my understanding on the pointer to multi-dimension array. Thanks a lot!

a[1] is not the first address of the 2-d array...it is the name of the 2-d array of rank 3x2...and since the name of an array is a synonym for the location of its initial element hence it contains the address of the second 2-d array of rank 3x2.

a[0] is this array -> {{'a', 'b'}, {'c', 'd'}, {'e', 'f'}}
a[1] is this array -> {{'g', 'h'}, {'i', 'j'}, {'k', 'l'}}
Yes your understanding is correct...
I'd suggest that you do those pointer arithmetic calculation using a pencil and paper...

Code:

char (*pa)[2] = &a[1][0];  /* original definition of pa */
pa points to -> {'g', 'h'}
pa[4] = *(pa + 4)  /* an array-and-index expression is equivalent to one written as a pointer and offset */
(pa + 4) points to -> {'o', 'p'}, and *(pa + 4) or pa[4] is that array via dereferencing

Since arrays are laid out sequentially...char 'r' is the 3rd element of the array "pa[4]" hence pa[4][3] == 'r'
If (*pa)[2] = &a[2][0] then you can get to the letter 'r' by...pa[0][5] or pa[2][1] or pa[1][3] or pa[3][-1]

Hope this helps...or have I confused you even more
For learning C I'd suggest sticking to "The C Programming Language" and "Expert C Programming"...those IMO are the 2 best books on the subject