Re: C++ help

[GP <EnderWiggenRules@aol.com>]

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In a message dated 3/14/01 3:42:47 PM Central Standard Time, 
tomdelduca@cs.com writes:


> dynamic_map = new char[ms_x] [ms_y];  (ms_x and ms_y passed into function)
>
> but this gives me a 'constant expression required' error.
> any help would be appreciated.


Ah, if only "new" was a bit more dynamic. Well, sorry to tell you this, but 
you must give a constant expression for "ms_x." In other words, you can't 
just say something like...

char *Buffer=new char[a][b][c];

...if a and b are variables. You HAVE to say something like this:

char *Buffer=new char[5][9][c];

Therefore, the rule about new is that only the last dimension (the "[c]") can 
be a variable. All preceding dimensions must be constants. There are a few 
ways around this. You could say:
#define a      5
#define b      9
...and then the first expression (char *Buffer=new char[a][b][c]) would be 
valid. This would also work:
const a=5, b=9;
These work because the compiler treats a and b as constants - not variables 
that can change during runtime. However, I bet you would prefer to 
dynamically (use variables) allocate memory for that array of yours. No 
problem. Here's how:
char *Buffer=new char[a*b*c];
OK, so you're not REALLY defining a three-dimensional array, but rather you 
are defining a one-dimensional array that can hold all the information the 3D 
array would hold. Just one thing to keep in mind. Say you wanted to access 
Buffer[x][y][z]. You can't just type that in because you do NOT have a 3D 
array. Instead, access this memory location:
Buffer[x*b*c+y*c+c];
TADA! The above command is just like accessing Buffer[x][y][z]. So, if you 
allocated your array like this:
dynamic_map = new char[ms_x*ms_y]; 
This would be perfectly legal. And, if you wanted to access 
dynamic_map[c][d], you would instead access dynamic_map[c*ms_y+d].

So, in practice, it is usually beneficial NOT to rely on multi-dimensional 
arrays. Instead, define it as a big one-dimensional array that can store all 
those values, and develop a way of organizing that data. The above method 
works well.

Hope this helps,
-Ender Wiggin
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