Thanks Corona688!
I fully understand the full array is actually a big 1-dimension array from your previous reply.
I tried:
Code:
printf("The value in a[1] is: %p\n", a[1]);
printf("The value in a[0] is: %p\n", a[0]);
printf("The value in a[1][0] is: %p\n", a[1][0]);
printf("The value in a[1][1] is: %p\n", a[1][1]);
Code:
Output:
The value in a[1] is: 0x7fff80c346a6
The value in a[0] is: 0x7fff80c346a0
The value in a[1][0] is: 0x7fff80c346a6
The value in a[1][1] is: 0x7fff80c346a8
The value in a[1][0][0] is: g
The value in a[1][0][1] is: h
first 4 printf() print addresses in fact, and I seem understand that:
Because a[1] is the first address of a two-way (3x2) array, so a[1] is the same as a[1][0], which in turn is the first address of two-elements array, so that (now value, not address) a[1][0][0] = g; a[1][0][1] = h.
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'}}
Quote:
Originally Posted by yifangt
Thanks Corona688! My confusion is: How is the connection with the pointer (*pa)[2]? Here the [2] is the last dimension of the array a[4][3][2]. I treat each "2-element array" as a box---a single unit. so that (*pa)[2] points a[4][3], each unit is an address of an array. Right?
Yes your understanding is correct...
Quote:
Originally Posted by yifangt
I thought pa[4][3] should point to an address of the array{'q', 'r'}, but it actually points to the element, char "r". I lost the connection here (*pa)[2] vs. pa[4][3].
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'
Quote:
Originally Posted by yifangt
Another confusion is the declaration of (*pa)[2] = &a[1][0] and (*ppa)[3][2] = &a[1]. The author is to emphasize the address of array and pointer. By array address only one anwer for char "r" which is a[2][2][1]. With pointer, you can have many ways(!?) What is the trick to connect two of them, say, if I declare (*pa)[2] = &a[2][0]?
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
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09-25-2013
