Dynamic array at C++

If I have something like:

    int **arr=new int *[3];
    arr[0]=new int [5];
    arr[2]=new int [12];
    arr[1]=new int [8];

and I make a new array:

    int **temp;
    temp=new int *[8];
    for (int i=0; i<3; i++)
      temp[i]=arr[i];

so the length of t[0], t[1], t[2], is same as arr[0], arr[1], arr[2]?
(I hope you understand what I mean...)

Thank you!!]

keskiverto (10402)

The answer is no. Not "same length". Pointers to same memory do they have.

cpcp (17)

You right...
But here is my problem:
here is the code:

int main()
{
	int n=3;
	int **m;
	//The initial array:
	m=new (nothrow) int *[n];
	for (int i=0;i<n;i++)
	{
			m[i]=new (nothrow) int[3];
			m[i][0]=i+1;
			m[i][1]=i+2;
m[i][2]=i+3;
}
for (int i=0;i<n;i++)
cout<<m[i][0]<<" "<<m[i][1]<<" "<<m[i][2]<<endl;
cout<<"----------------------------------------------\n";
int **t; //The temporary array:
t=new (nothrow) int *[n*2];
for (int i=0; i<n; i++)
{
	t[i]=m[i];
	t[i+n]=new int[3];

	cout<<t[i][0]<<" @ "<<t[i][1]<<" @ "<<t[i][2]<<endl;
}
	cout<<"\n\n\n";
for (int i=n; i<n*2; i++)
{
for (int j=0; j<n; j++)
t[i][j]=j+i;
cout<<t[i][0]<<" % "<<t[i][1]<<" % "<<t[i][2]<<endl;
}
cout<<"\n\n\n";
for (int i=0;i<n;i++)
delete []m[i];
delete []m;
m=t;  //#####//

//Printing the final array:
for (int i=0;i<n*2;i++)
cout<<m[i][0]<<" "<<m[i][1]<<" "<<m[i][2]<<endl;

for (int i=0;i<n;i++)
delete []t[i];
delete []t;
return 0;
}

And here is the output:

1 2 3
2 3 4
3 4 5

1 @ 2 @ 3
2 @ 3 @ 4
3 @ 4 @ 5



3 % 4 % 5
4 % 5 % 6
5 % 6 % 7



0 2 3
153640976 3 4
153640992 4 5
3 4 5
4 5 6
5 6 7

you see the bold lines?

1
2
3

0 2 3
153640976 3 4
153640992 4 5

why I got those numbers?
I just tried to copy the array (t -> m).

Thank you!

writetonsharma (1461)

You are trying to access already deleted items. You deleted the first three indexes and iterating them will give you unpredictable results.

cpcp (17)

I fix the code, and still, the array did not copied well:

#include <iostream>

using std::cin;
using std::cout;
using std::endl;
using std::nothrow;
using std::cerr;

int main()
{
	int n=3;
	int **m;
	//The initial array:
	m=new (nothrow) int *[n];
	for (int i=0;i<n;i++)
	{
			m[i]=new (nothrow) int[3];
			m[i][0]=i+1;
			m[i][1]=i+2;
			m[i][2]=i+3;
	}
	for (int i=0;i<n;i++)
		cout<<m[i][0]<<" "<<m[i][1]<<" "<<m[i][2]<<" "<<endl;
	cout<<"----------------------------------------------\n";

	int **t; //The temporary array:
	t=new (nothrow) int *[n*2];
	for (int i=0; i<n; i++)
	{
		t[i]=m[i];
		t[i+n]=new int[3];
		cout<<t[i][0]<<" @ "<<t[i][1]<<" @ "<<t[i][2]<<endl;
	}
	cout<<"\n\n\n";
	for (int i=n; i<n*2; i++)
	{
		for (int j=0; j<n; j++)
			t[i][j]=j+i;
		cout<<t[i][0]<<" % "<<t[i][1]<<" % "<<t[i][2]<<endl;
	}

	cout<<"\n\n\n";
	for (int i=0;i<n;i++)
		delete []m[i];
	delete []m;
	m=new int *[n*2];
	for (int i=0; i<n*2; i++)
	{
		m[i]=t[i];
	}

	//Printing the final array:
	for (int i=0;i<n*2;i++)
		cout<<m[i][0]<<" "<<m[i][1]<<" "<<m[i][2]<<" "<<endl;

	for (int i=0;i<n;i++)
		delete []t[i];
	delete []t;
	return 0;
}

Here is the output:


1 2 3 
2 3 4 
3 4 5

1 @ 2 @ 3
2 @ 3 @ 4
3 @ 4 @ 5



3 % 4 % 5
4 % 5 % 6
5 % 6 % 7



0 2 3 
136372240 3 4 
136372256 4 5 
3 4 5 
4 5 6 
5 6 7

Can you tell me where I wrong please?

Thank you!

Last edited on

MiiNiPaa (8886)

Line 17: You allocated some memory and made m[i] point to it
Line 30: You made t[i] point to same memory area as m[i]
Line 44: You deleting memory both t[i] and m[i] pointing to; both pointers now invalid.
Line 49: you are copying invalid pointer from t[i] to new m[i] array.

Last edited on

cpcp (17)

Thank you.
Now I'm understand it more clearly....

But what I can do to avoid:

Line 44: You deleting memory both t[i] and m[i] pointing to; both pointers now invalid.
Line 49: you are copying invalid pointer from t[i] to new m[i] array.

At the next level I going to build a function that do it (add more lines to a dynamic array) so I have to delete one array in the function (and this why I delete m at the middle of the code).

Thank you :-)

MiiNiPaa (8886)

Also in your code if you do

1
2
3

m[1][1] = 5;
t[1][1] = 3;
std::cout << m[1][1];

it will give you "3" because it is the same memory.
To get rid of all such behavior:

1
2
3

t[i] = new int[n];
for(size_t j(0); j < n; ++j)
    t[i][j] = m[i][j];

instead of t[i] = m[i];

Last edited on

Topic archived. No new replies allowed.

C++

Forum

Dynamic array at C++