Hi,

I have a pointer to a 2D array and I declared it as follows

T (*element_) [ ];


and I allocate my array as follows, (in the constructor ofcourse)

T (*element_)[n];
element_ = new T [m][n];


I'am trying to get a member function return the base address of the individual rows. But I'm in a kind of fix when attempting to declare a function which will do that.

Can someone suggest me how I declare a function, which will return the address of each rows ?

Thanks

I'm not familiar w/ your syntax, but don't forget that a pointer IS the address... just print it
ex:
this gives me:
which are the addresses of the rows.

or I suppose you could do:
but (I don't think) there's any point of that ( no pun intended

EDIT:
I guess you don't want to print them (I'm _really_ tired), but this should still illustrate how to get what you want (do something other than cout them )
sorry for getting off track.

hi,
in your declaration T (*element_)[n]; there are several errors:
1. the size , the value of n, must be known at the compiling time, because you decalre a static arrays, if you want a dynamic arrays use news and a pointer to pointer like char **tata;
2. the parenthesis after T , decalre T as a function , use instead bracket [.


This is an example:
#include<iostream.h>


void func1( int **ptr, int row1, int col1){
cout << "give "<< row1 <<" times " << col1 << "numbers"<<endl;
for (int i=0; i < row1; i++)
for (int j=0; j < col1; j++)
cin >> ptr[i][j];
}
int main(){
int row,col,**x; // for dynamic allocation
cout << "enter nb of rows "; cin >> row;
cout << "enter nb of cols "; cin >> col;

x = new (int*)[row];
for(int i=0;i<row;i++)
x[i]= new int[col];
// call a function

func1(x,row,col);
// here you get the array x filled in the funct1
// the address of the rows are x[0], x[1]...
for(int i=0;i<row;i++)
cout << "x[ "<< i << "] = " << x[i] << endl;

return(0);
}
/*this is a test ;

enter nb of rows 3
enter nb of cols 2
give 3 times 2numbers
11 22
33 44
55 66
x[ 0] = 0x804a118
x[ 1] = 0x804a128
x[ 2] = 0x804a138
*/

I think that you are printing out the address of the the first element in each colum here, not the addresses of each row.

it's like doing this:
where you are printing the value of memory that (x+i) holds.

Of course, this may be what dhanakom wants... it just depends which one you call row/colum

Hi,

All the replies suggest a different way of declaring a pointer to the 2D array.

But the following works jus fine

T (*ptr)[n];
//meaning ptr will hold the address of an array(base address) of n elements of type T

and when I again say
ptr = new T [m][n];

I mean ptr will hold the base address of an array of 'm' elements each of which is an array of 'n' elements. Note that the second 'n' becomes a part of the type.I have a running program with this declaration. This is exactly what devoyage code does, but for few changes in the syntax.

But suppose u want to declare a function which will return the address of each row, how will u declare the function prototype

should it be T (*) [] fun(parm1,param2,param3)

this apparently means this function will return the address of an array(base address) whose elements are of type T. But the above declaration gives a compilation error.

what might be wrong here ??


Thanks,

Except for the fact that n must evaluate to a constant at compile time, i.e. not dynamic. ( I still don't understand that other part

Just like I said in your other post:
This is quite similar to what you have, just conforming to compiler needs a l ittle more I guess.

You can't return all of the addresses, because you can't 'return' arrays, only the address of the first element. (unless you want to throw them into some kind of structure I guess)

Sorry if I'm not understanding your questions. I hope this gives you some ideas at least.

-Devoyage

hi,
to Devoyage,
the addresses are printed bye the program, you can test it !!!
In fact it is not the same as you code !!
in your case x = new (int*)[row];

x is an array for row (5 ) elements each one contains an address, the address of the first is the same as x , the address of the second is the next 4bytes, and so on,
that is :
0x804a040 location of x same as location of x[0]
0x804a044 location of x[1]
0x804a048 location of x[2] not the contains of it
0x804a04c
0x804a050

now try to print x[i] and you get the same thing as mine

best regards

I couldn't agree more, that is what I was trying to say. Maybe we were misunderstanding eachother or something.

Now, here's the output of a new and inproved tester ...
Like I said before, it depends on which one you call row,colum.

I was using the convention var[ROW][COLUM], and by the looks of your declerations, I thought you were too.

to print the address of the first element in each row, I used 'x+i', which is what he asked. You were printing the address of the first element of each colum, namely 'x[i]', the contents of 'x + i'.