FILE I/O problem please help

[mayankladoia]

Following is the code which i am running on arr[2000][2000] array:

ifstream inf;
inf.open("ABC1. txt");

for(i=0;i<x;i++ )
{
for(j=0;j<x;j++ )
{
inf>>arr[i][j];
if(arr[i][j]==0)
{
ind[i][j]=0;
}
else
{
ind[i][j]=1;
}
}
}

after running this code arr[][] contains only 0 in all cells ie arr[][]= 0 is
0 0 0 0 0 0 0 0 ............ 2000 times
0 . . . ............... .....
0 . . . ............... .....
0 ............... ..........
0 ............... .........
.
.
.
2000 times

This only happens if size of x is 2000 for x=1000 same code is running fine.

Can you suggest solution for the same.....
Thanks in advance

[weaknessforcats]

I suspect your array is too large for the stack.

A 2000x2000 array is 4 million elements.

If the array is char that's a 4MB stack array. If the array is int ons 32 bit machine that's a 16MB array.

A big stack frame is 8K.

You should be able to allocate the array on the heap and it should work OK at 2000x2000.

[mayankladoia]

This is the beginning of the code ... It is like an input text file is given which is an array of n X n i.e a 2D array value of n is 2000. This text file is opened by following lines

ifstream inf;
inf.open("ABC1. txt");

And then the above given for loop runs on this code. when i am trying to display arr array by adding this line after inf>>arr[i][j] in the above code
cout<<arr[i][j]<<"\n";

It is only displaying 0 for more than 2000 elements which clearly indicates that it is unable to read data from text file for large input
I dont know how to use heap for this please tell me in case u have any idea

[weaknessforcats]

Allocate on the heap by:

int (*arr)[2000] = new int[2000][2000];

assuming your array is an array of int.

The code aboive is exactly the same as:

int arr[2000][2000];

With the heap array ytou need to delete it when you are done with it:

delete [] arr;

See if this makes a difference. Be sure to post again and tell me what happened.

BTW: There is a C/C++ Insight paper called Arrays Revealed that you might read.