Thread: Hello all and pointers to multi-dimensional arrays

First and foremost, Hi
I'm new here and since only recently I started to learn C++, I think I'll stick around

I've been following the excellent tutorial here ate cprogramming as part of my learning, changing a lot of code to better see if I really understood each subject. On one of those changes I got stalled.

Lesson 8 teaches arrays. Nothing wrong here, only when I was doing my own experiments, I applied what I had learned in lesson 6 - Pointers. Great lesson, btw! I finally was able to understand the buggers. Anyway... I tried to pass a pointer into a multidimesional array. Like this

Code:

    char mycharr[8][8];
    char *ptrarr;
    ptrarr = mycharr;

And it didn't work. I get a compile error like this:

Code:

assignment to `char *' from `char (*)[8]'

It works fine for normal arrays, but I still couldn't figure it out how to use it on multi-dimensional ones even after messing with the above code.

I use mingW that comes with Blodsheed's Dev-C++ IDE. Don't ask me compiler versions... I'm not even near that stuff

I believe when you use a pointer it will be treated as a 1-dimensional array, even if you point it at a 2-dimensional.

Maybe if you made teh pointer a 2D Array like so:

Code:

    char mycharr[8][8];
    char **ptrarr;
    ptrarr = mycharr;

(i think thats how u do it, im not positive though)

Originally posted by Okiesmokie
Maybe if you made teh pointer a 2D Array like so:

Code:

    char mycharr[8][8];
    char **ptrarr;
    ptrarr = mycharr;

(i think thats how u do it, im not positive though)

I doubt it, since you wouldn't know if the array you point at is a 8x8, 4x16 or 2x32 array. This would cause trouble when indexing.

A possible solution would be:
char* ptrarr[8];

Or simply treat the 2D array as a 1D array:
char* ptrarr;
ptrarr[Y * 8 + X] = 'A';
Var = ptrarr[y * 8 + X];

Originally posted by Magos

Or simply treat the 2D array as a 1D array:
char* ptrarr;
ptrarr[Y * 8 + X] = 'A';
Var = ptrarr[y * 8 + X];

Err... I'm not sure I understood that code. I mean the part about placing the * immediatly after the keyword.
Nevertheless the problem here is that the compiler gives an error not when trying to use the pointer, but before that, when defining it. That is,

Code:

char mycharr[8][8];
char *ptrarr;                //compiles ok... obviously :cool: 
ptrarr = mycharr;         //complains here

Hmm, I made some tests on this one and I realized that some of my solutions didn't work so well. This is an example that works however:

Code:

int main()
{
   char mycharr[8][8]={0};
   mycharr[0][0]='A';
   mycharr[0][2]='B';
   mycharr[2][0]='C';
   mycharr[3][5]='D';

   char *ptrarr;
   ptrarr = *mycharr;

   printf("%c %c %c %c", ptrarr[0*8+0], ptrarr[0*8+2], ptrarr[2*8+0], ptrarr[3*8+5]);
   getch();
   return 0;
}

Since mycharr is an array of arrays, you must set the pointer ptrarr like this:
ptrarr = *mycharr;
(The adress of the first array)
instead of what I said in the previous post:
ptrarr = mycharr;
(The adress of the array of arrays)
The above example also shows how you can simulate a 2D array using a 1D array (using X*Width+Y, or X*8+Y).
Hope I was to some help .

hmm what about this:

Code:

    char mycharr[8][8];
    char **ptrarr;
    ptrarr = new char[8][8];
    ptrarr = mycharr;

??

Originally posted by Magos

Code:

int main()
{
   char mycharr[8][8]={0};
   mycharr[0][0]='A';
   mycharr[0][2]='B';
   mycharr[2][0]='C';
   mycharr[3][5]='D';

   char *ptrarr;
   ptrarr = *mycharr;

   printf("%c %c %c %c", ptrarr[0*8+0], ptrarr[0*8+2], ptrarr[2*8+0], ptrarr[3*8+5]);
   getch();
   return 0;
}

[SNIP]
The above example also shows how you can simulate a 2D array using a 1D array (using X*Width+Y, or X*8+Y).
Hope I was to some help . [/B]

Oh, Yes you were!
It compile nicely and also helped me out understand it better. Thanks a lot

Originally posted by Okiesmokie
hmm what about this:

Code:

    char mycharr[8][8];
    char **ptrarr;
    ptrarr = new char[8][8];
    ptrarr = mycharr;

??

Sorry to dissapoint you, but:

Code:

int main()
{
   char mycharr[8][8];
   char **ptrarr;
   ptrarr = new char[8][8];    //Line 8
   ptrarr = mycharr;           //Line 9
   getch();
   delete[] ptrarr;
   return 0;
}

Error:  pointer01.cpp(8,12):Cannot convert 'char ( *)[8]' to 'char * *'
Error:  pointer01.cpp(9,12):Cannot convert 'char ( *)[8]' to 'char * *'

Perhaps you should try out your code before you post it? I know I should

whoa! that's an helluva of information there.
I didn't realize the question had such implications... too much info for this poor ol' noob

Anyway, I'll sure take a close look at that tutorial. 12 pages on pointers just can't be wrong

I'll also remember to search for answers also in C when getting stalled.

Thanks

[I tried to pass a pointer into a multidimesional array. Like this

Code:

    char mycharr[8][8];
    char *ptrarr;
    ptrarr = mycharr;

I think /.....

Code:

    char mycharr[8][8];
    char *ptrarr;
    ptrarr = mycharr; // here you need to overload " = "

Does it needs some overloading...?

Did it compile... at the end...