# Quick Help!

• 12-25-2002
Ice_Black
Quick Help!
I have a int varible: int a = 1234

then i want to return each digit seperately like this.:
1
2
3
4

any how?
• 12-25-2002
RoD
This would count out 1,234 in order...

Code:

```int main() { for (int i = 0; i <=1234; i++) { cout<<i<<endl; } return 0; }```
this would show your a= 1234 in reverse ordeR:

Code:

``` int main() { int a = 1234; while(a > 0) { cout<<a<<endl; a--; } return 0; }```
• 12-25-2002
Jaguar
If you are using BC++, try itoa() in stdlib.h
• 12-25-2002
Ice_Black
I dont want it like that..

i dont want the number to decrease.

int a = 976724;

so it would return, 9,7,6,7,2,4;

so i would be able to add 9 + 7 + 6 + 7 + 2 + 4 from int a.
• 12-25-2002
RoD
i don't think you can do that.
• 12-25-2002
Jaguar
Try something like

int i=10;
int a=123456;
int d;

a%i must be 6 --to be kept (may be in an array)
d=a/i; // d becomes 12345
a=d;
a%i must be 5 --to be kept

do the same in some routines.
sorry, I did not write program here, just an idea.
• 12-25-2002
RoD
the idea of more variables would work but hes saying he wants one variable, and wants to add(i think it was add) the numbers in it.

a = 12

(1 + 2)

a would = 3;
• 12-25-2002
Eibro
One way I can figure
Code:

```  int num = 1234567;   int n[7];   n[0] = num - (num/10)*10;   n[1] = (num - (num/100)*100 - n[0])/10;   n[2] = (num - (num/1000)*1000 - n[1])/100;   n[3] = (num - (num/10000)*10000 - n[2])/1000;```
Now; put that in a recursive function, because I sure as hell can't.
• 12-25-2002
PJYelton
You could do it with one more intermediate variable.
Code:

```int a=12345; int num=0; while (!a) {   num+=a%10;   a/=10; } a=num;```
• 12-25-2002
Eibro
Quote:

Originally posted by PJYelton
You could do it with one more intermediate variable.
Code:

```int a=12345; int num=0; while (!a) {   num+=a%10;   a/=10; } a=num;```

I believe that should be while (a), so

Code:

```int get_sum(int num) {   int sum =0;   while (num)   {       sum+=num%10;       num/=10;   }   return sum; }```
Nice solution though.
• 12-26-2002
PJYelton
Whoops, alright who put that '!' operator in my code?? ;) :D
• 12-26-2002
workable version, uncompiled

Code:

```#include <iostream> using namespace std; int main() {   long input;   int array[10];   int index = -1;     cout << "enter an integer of value less than 10,000,000 << endl;   cin > input;   while(input)   {     array[++index] = input % 10;     input  /= 10;   }     for( ; index > -1; --index)   {       cout << array[index] << endl;   }   return 0; }```
• 12-26-2002
Quote:

workable version, uncompiled

Code:

```#include <iostream> using namespace std; int main() {   long input;   int array[10];   int index = -1;     cout << "enter an integer of value less than 10,000,000" << endl;   cin >> input;   while(input)   {     array[++index] = input % 10;     input  /= 10;   }     for( ; index > -1; --index)   {       cout << array[index] << endl;   }   return 0; }```

2 Errors were in the code I have added ,mentioned in red color..

and I have write a program ... which also requires to seperate each input digit... for adding them with each.. other...It could help too..
Code:

```include <iostream.h> #include <conio.h> #include <stdlib.h> #include <string.h>  int main(void)   {   int leng,res=0; //leng for length of input number.   long int number,divider=1,remainder; //   char *value; //value will be recieved in string...   clrscr();   cout<<"Enter any long integer: ";   cin>>value;   leng=strlen(value); //find the length of the string.   number = atol(value);   for (int i = 0; i<leng-1; i++)     {     divider = divider * 10;     }     res = 0; //note result is intitiazed to zero.   for (i = 0; i<leng; i++)     {     res +=  number / divider;     number = number % divider;     divider = divider / 10;     }   cout<<"Result is:\t"<<res<<endl;   getch();   return 0;   }```
• 01-10-2003
Ice_Black
...
Thanks you...