0

this code is working for a counter of 6 bit and i need a counter of 48 bit ... if u have time kindly try it..... i try alot but not succeed

```
#include<stdio.h>
#include<stdlib.h>
#include<float.h>
#include<math.h>
#include <time.h>
#include <cstdlib>
#include <iostream>
#include<stdint.h>
#include<conio.h>
int main()
{
int a[64][6];
int i,j,k,d,c,n;
for(k=0;k<64;k++) //6 digit binary so 2^6 =64 is no: of combinations
{
for(j=0;j<6;j++)
{
a[k][j]=0;
}
}
for(i=0;i<64;i++)
{
n=i;
for(j=5;j>=0;j--)
{
while(n!=0)
{
a[i][j]=n%2;
j--;
n=n/2;
}
}
}
for(k=0;k<64;k++)
{
for(j=0;j<6;j++)
{
printf("%d",a[k][j]);
}
printf(" .................%d\n",k);
}
//printf("\n **** KHATAM SHo ****\n");
getch();
return 0;
}
help please
```