[dan_ram]

why is this not possible? i want to obtain the no. of rows and then allocate memory from the heap. if there's no alteration that can be done to this, what is the alternative?

#include<iostream.h>
int main(){
   int n;
   cout<<"enter no. of columns/rows of square matrix"<<endl;
   cin>>n;
   int *p=new int[n][n];
   return 0;
} 

[KYA]

This is an example in C (see code comments):

#include<iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;

int main()
{
   int n = 0;
   cout<<"enter no. of columns/rows of square matrix"<<endl;
   cin>>n;
   int **p;
   p = (int**)malloc(n*sizeof(int*)); //makes p[n], where n is the number of pointers, C++ requires cast
   
   for(int i = 0; i < n; i++)
   {
		   p[i] = (int*)malloc(n*sizeof(int)); //for each p[n]we make the second dimension (# of ints 'n')
   }
   
   int a  = 0;
   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   p[i][j] = a++; //asign some values, never take actions on uninitialized dynamically allocated arrays 
		   }
   }
   
   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   cout << p[i][j] << " "; //print them out
		   }
		   cout << "\n";
   }
   
   for (int i = 0; i < n; i++)
   {
		free((int*)p[i]); //free up the array of pointers, must free each one otherwise memory will leak
   }

   free((int**)p); //free the handle to the multi dimension array
   system("pause");
   return 0;
}

[dan_ram]

whats a cast? I'm not supposed to use C nor am i aware of it.
can't new and delete be used?

[KYA]

It can. You just can't do it like you did. My snippet was to illustrate how to do it in C++ by giving an example in C. Not subtle enough?

//In C++ Exclusively
#include<iostream>
using namespace std;

int main()
{
   int n = 0;
   cout<<"enter no. of columns/rows of square matrix"<<endl;
   cin>>n;
   int **p;
   p = new int*[n]; //make an array of ints, p points to them

   for(int i = 0; i< n; i++)
   {
	   p[i] = new int[n]; //each pointer now points to 'n' amount of ints
   }

   int a  = 0;
   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   p[i][j] = a++; //assign some values, never take actions on uninitialized dynamically allocated arrays 
		   }
   }

   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   cout << p[i][j] << " "; //print them out
		   }
		   cout << "\n";
   }

   for (int i = 0; i < n; i++)
   {
		delete((int*)p[i]); //free up the array of pointers, must delete each one otherwise memory will leak
   }
   
   delete [] p; //delete the handle (its an array so use brackets)

	
   system("pause");
   return 0;
}

If you don't know what a typecast is why are you making a program with a dynamically allocated multi dimensional array? :Just curious...

This post has been edited by KYA: 25 December 2008 - 10:01 AM

[baavgai]

I'll have to preface this by saying that I loathe multidimentional arrays. While they're often simple conceptually, the implementation is rarely worth it.

To dynamically allocation a 2d array, you essentially have to make a "jagged" array

int n = 3;

// First, the 1d array.
int *p1d;
p1d = new int[n];
// Nice, simple, conceptually clean.  You have pointer to a block of n ints

// 2d is more difficult
int **p2d;
// nice syntax, huh?
// This first bit isn't so bad
*p2d = new int[n];
// you've allocated the first dimension
// even though it says int, it's really a list of pointers.

// we still need to allocate the actual storage
// and we need to do that for each n
for(int i=0; i<n; i++) {
	p2d[i] = new int[n];
}

That's the most basic example. You'll probably want to read more online.

Good luck.

Edit: oops, KYA beat me. Interesting, both syntaxes work. C++ is fun for that.

This post has been edited by baavgai: 25 December 2008 - 10:15 AM

[dan_ram]

thnx a lot kya and baavgai..
hey..i wasnt aware of the word "typecast"
i was just trying it out with the knowledge of 1D array..
hey 1 doubt...wen you declare a pointer, it has to be defined then and there right?
how can you give
int*p;
p=.....; ????????

This post has been edited by dan_ram: 25 December 2008 - 11:17 AM

[KYA]

Nope, although its good practice to do so immediately if not immediately following:

int* p; //perfectly legal

However if you aren't going to use it for a while, initalizing it to zero (effectively null) is also good:

int* p = 0;

[Locke]

KYA, on 25 Dec, 2008 - 08:41 AM, said:

This is an example in C (see code comments):

#include<iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;

int main()
{
   int n = 0;
   cout<<"enter no. of columns/rows of square matrix"<<endl;
   cin>>n;
   int **p;
   p = (int**)malloc(n*sizeof(int*)); //makes p[n], where n is the number of pointers, C++ requires cast
   
   for(int i = 0; i < n; i++)
   {
		   p[i] = (int*)malloc(n*sizeof(int)); //for each p[n]we make the second dimension (# of ints 'n')
   }
   
   int a  = 0;
   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   p[i][j] = a++; //asign some values, never take actions on uninitialized dynamically allocated arrays 
		   }
   }
   
   for (int i = 0; i < n; i++)
   {
		   for(int j = 0; j < n; j++)
		   {
				   cout << p[i][j] << " "; //print them out
		   }
		   cout << "\n";
   }
   
   for (int i = 0; i < n; i++)
   {
		free((int*)p[i]); //free up the array of pointers, must free each one otherwise memory will leak
   }

   free((int**)p); //free the handle to the multi dimension array
   system("pause");
   return 0;
}

Now hold on a second there KYA...there's no such thing as iostream in C. (at least as far as I'm aware.) Nor is there cin or cout...again, as far as I'm aware. I think yuo're a bit confused.

stdio.h and printf(), scanf() perhaps?
________________________________________________

@dan_ram:

When you declare a pointer, it's perfectly legal like KYA said to just do this -> int* p;, however, when you initialize it to something...you can initialize it 2 different ways.

int *p; // again...perfectly legal

int a = 5; // regular int variable

p = &a; // make the pointer 'p' point to the variable 'a'

*p = 10; // make the value pointed at by 'p' equal to 10

cout << a; // I believe this will be 10...however, I'm not COMPLETELY sure.

Hope this helps!

This post has been edited by Locke37: 25 December 2008 - 12:07 PM

[baavgai]

dan_ram, on 25 Dec, 2008 - 12:16 PM, said:

int*p;
p=.....; ????????

This is entirely correct. You have no clue what *p contains. I'd prefer this syntax.

int *p = NULL;

What you've done is declared a pointer and then assigned null to that pointer. However, in most cases, you can declare a value without assigning it anything. While I generally prefer to put a value in, pointers are a little confusing.

Here, it's clear that the address is being assigned to pointer p:

int n = 3;
int *p;
p = new int[n];  // setting the address

Here, it's doing some tricks:

int n = 3;
int *p = new int[n];

Put another way,

int n = 3;

// you can do this
int *p = &n;

// you can do this
int *p;
p = &n;

// you can't do this
int *p;
*p = &n;

Confusing, no?

[KYA]

Whoops, I should have said: C style functions, barring I/O processes.