I need to find the smallest number formed from the same digits of a given natural number n, using arrays.

#include <fstream>
#include <math.h>

int *list;
bool arrange(int length){

  for(int i=0;i<length;i++)



  for(int i=0;i<length;i++)cout<<"my new list contains "<<list [i]<<endl;

return true;
bool file(char* Filename, int length)


	ifstream file;

	list=new int[length];
  for(int i=0;i<length;i++) 
	cout<<"the list contain: "<<list[i]<<endl;


return true;

Say you have the number: 251 The smallest number from those digits will be: 125 Therefore: digit1 = smallest number, digit2 = next smallest, etc.

Seperate the number into individual digits using an array. Find the smallest value and put this in the first index of a new array. Repeat this, removing the number you use each time, until you have no numbers left in the original array. Put all of the new array's values into a string and convert that to an int. This is your new number!

EDIT: You may want to use lists not an array.
EDIT EDIT: Oops got beaten too it. Ah well :)

I've found a code that works very well:

using namespace std;
int main()
    int n,a[10],i,j;
    for(i=0;i<=9;i++) a[i]=0;
    while(a[i]==0) i++;
    return 0;

I think it's still difficult for me to understand why for(i=0;i<=9;i++) a[i]=0;
before cin>>n and what is the significance of j.
Otherwise, the code looks like you are trying to explain.

Next time, please press the code button before you paste your code.

for (i=0; i<=9; i++){}

Ever done a while loop before? Consider a loop which is while (i<=9){} . At the end of the loop, you would have i++; to increment the value of i .
Basically (in a for loop): i=0 makes a new variable called i which is set to 0 . i<=9 means that we keep looping while i is less than or equal to 9. i++ means that we increment the value of i once each loop.

I didn't understand why for(...) at the beginning, a has to be 0.

I didn't understand why for(...) at the beginning, a has to be 0.

Who says it has to be 0? Test it. First thing in the program just print out the array.