Two dimension array as Function Parameter

[ysaacb]

Hi Everibody

How can I a pass a two dimension array asa function parameter?

Regards

Ysaac

[Ken Johnson]

Pass the name of the array (it's a pointer to the array). Note, however, that the function does not know the dimensions.

e.g.
[code]
void myfunc (char *p)
{
..
}

{
char a[10][5];

myfunction (a);
}
[/code]

should work ok.

Ken

[ysaacb]

I'm trying to do something like this i.e. make the array point to the adress of a

void myfunc (char *p)
{

char BUFFER[7][32];
buffer=a;
..
}

{
char a[10][5];

Regards
Ysaac

myfunction (a);
}

[Ttelmah]

You are not goint to get very far with this.
The declaration:
char BUFFER[7][32];

Actually _creates_the array. So you are immediately using 224 bytes of storage. An array name is a _constant_, so you can't then point this to another external array. You need to declare 'buffer' as:
char *buffer;

Which is then a variable that can be used to point to an array.
You then seem to want to use different dimensions inside the function to outside. A sure way of destroying memory data....
What you can do, is pass a pointer to a function, for use as an array, but you have to specify _one_ dimension.

[Wayne_]

A couple of things. In your code in the function you are doing :-
buffer = a;
Your value is actually p as defined in the function header so
buffer=p;

char BUFFER[7][32]; is not required unless you are using BUFFER for something else as C is case sensitive.

So you are almost there :-
void myfunc(char *p)
{
char *buffer = p;

buffer[0][0] = 'a';
buffer[0][1] = 'b';
strcpy(buffer[1], "Hello");
}

void main()
{
char a[10][5];
myfunction(1);
}

Should work and do what you want.

Actually unless you need to keep a reference to the start of the array (p) or you are not actually going to modify the pointer (p) you can do away with buffer and just do
p[0][0] = 'a';
etc...

[Guest]

Dimensions restricted by 5x5