How do I seperate integers?

**Guti14** · 12-21-2003

Hi,

I'm doing this little program at the moment where I have to imput a 5-digit integer, seperate the number into its individual digits and prints the digits seperated from one another by three spaces each.
It gives me a hint that I should use the division and modulus operators.

Iv'e started like this, but really am clueless how to tackle this, can anyone give me a few tips and/or pointers

thanks for any help

Code:

int main(int argc, char* argv[])
{
   int number;

   cout << "Enter a five digit integer: ";
   cin >> number;

   number = number / 10;
   cout << number;

   number = number / 100;
   cout << number;

   number = number / 1000;
   cout << number;

   number = number / 10000;
   cout << number;

   getch();
   return 0;
}

**silk.odyssey** · 12-21-2003

Here's how I would do it

If 51236 is entered,

Code:

 
51236 / 10000 = 5 rem 1236 
 1236 /  1000 = 1 rem 236 
  236 /   100 = 2 rem 36 
   36 /    10 = 3 rem 6 
    6 /     1 = 6

**Guti14** · 12-21-2003

yes, but the input is not pre-determined. and i have never comes across "rem" before yet.

**Codeplug** · 12-21-2003

Modulus Operator: %

gg

**silk.odyssey** · 12-21-2003

Guti14,

rem just meant remainder.

I know the input is not predetermined, I was just suggesting a way you could write the program.

51236 / 10000 = 5 rem 1236

51236 divided by 10000 is 5 with 1236 remaining. So we know that the first number is 5.

1236 / 1000 = 1 rem 236

1236 divided by 1000 gives us 1 with 236 remaining. So we know 1 is the second number.

236 / 100 = 2 rem 36
36 / 10 = 3 rem 6
6 / 1 = 6

Same idea for the other numbers. It's a simple algorithm and it should work for any 5 digit number.

**Zach L.** · 12-21-2003

For an arbitrary number of digits, you need only compute the number of decimal digits to use in a loop. The number of digits will be floor(log10(n)) + 1.

**grib** · 12-21-2003

You don't need to know the number of digits to print it out backwards, this is probably giving too much help but.

Code:

    std::cout << n % 10;
    while(n /= 10) std::cout << "   " << (n % 10);
    std::cout << std::endl;

**Zach L.** · 12-22-2003

^ True, but to print it out forward you probably should, but it could be done otherwise:

Code:

unsigned reverse(unsigned x)
{
  unsigned t = 0;

  do
  {
    t *= 10;
    t += x % 10;
    x /= 10;
  } while(x);

  return t;
}

And then grib's code would work.

**SirCrono6** · 12-22-2003

silk.odyssey,

Actually, those aren't reamainders. 51236 / 10000 = 5.1236
Note the decimal. So they aren't reamainders, but decimals. Use a calculator if you don't believe me.

-SirCrono6

**grib** · 12-22-2003

Sir Crono, we aren't using a calculator, we're using C++, the result of int/int is another int (truncated (a/b)*b+a%b==a).

Zach, I think your right and the op wanted it forwards, hmmm. Must come up with goofier way do to this.

Code:

void f(unsigned n) {if(n) {f(n/10); std::cout << (n % 10) << "   ";}}

**SirCrono6** · 12-22-2003

Yeah, I know. I guess I was just thinking math - math.

Int, integer, whole number. Float, float, decimal. Char, character, letter...

-SirCrono6

**Zach L.** · 12-22-2003

Originally posted by grib

Code:

void f(unsigned n) {if(n) {f(n/10); std::cout << (n % 10) << " ";}}

Nice.

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