Can someone please tell me if there is anything incorrect with this initialization:
static unsigned char* var[1]
var[0] = new unsigned char[ 800 * 600 ]
Is this creating a 2D array at var[0] ? Is this even valid ?
5 Answers
It's creating a single array with 480,000 elements (aka 800*600) and storing it in var[0]. It is not creating a 2D array. It's valid as long as var[0] is an unsigned char*.
5 Comments
c0d3rz
Does that mean, it is over-writing var[1], var[2] etc.?
Edward Loper
No -- if
var[0] is an unsigned char* then var must be an unsigned char** or a std::vector<unsigned char*> or some other such type. Assigning to var[0] will not affect var[1] or any other element of var.
Bo Persson
As
var only has one element, var[0] is all there is!
betabandido
@c0d3rz
var[1] does not exist since your array contains only one element.c0d3rz
Thanks Edward for the reply. It gives me a lot more understanding. I'm relatively new to C++. Sorry for re-editing the post. I wish I could accept more than one correct answer ...
It is creating a 1-d array of length (800*600) at var[0]. You'll be able to access its elements with:
var[0][0]
var[0][1]
...
var[0][800*600-1]
Comments
I think you want something more like (although I may be interrupting the question wrong);
static unsigned char* var = new char*[800]
for (int i =0; i < 800; i++)
var[i] = new char[600]
That will give you a 800x600 2d character array.
Comments
Your code is not correct, since var[0] is not a pointer anymore.
You may want to do something like:
static unsigned char* var;
var = new unsigned char[800 * 600];
That does not create a 2D array. It is just a 1D array. Having said that, you can use it as a 2D array if you compute the offset yourself. For instance to access position (row, col) you could do:
var[row * 600 + col]
If you really want a "true" 2D array, you will need to use for instance:
static unsigned char** var;
var = new unsigned char*[600];
for (int i = 0; i < 600; i++)
var[i] = new unsigned char[800];
Of course, if you do not need to dynamically specify the array size at runtime, you can just use a statically allocated array:
static unsigned char var[600][800];
BTW, I assume 600 is the number of rows and 800 the number of cols (like in a matrix). That is why I always use 600 in the first index. Otherwise, just swap the values.
4 Comments
K-ballo
var[0] is of type unsigned char*.betabandido
@K-ballo the OP changed the code. Before it was just
unsigned char* var;Bo Persson
Or you could do
unsigned char var[600][800]; and save yourself a lot of trouble.betabandido
@BoPersson That's true, but I guess the OP would not use dynamic allocation if his/her problem could be solved with a statically allocated array.
Yes, it's quite incorrect to use new[]. What you're looking for is std::vector<unsigned int> var(800 * 600);.
Comments
varand the type ofvar[0]are not the same thing.