Thread: Need some suggestions/Guidance

Hello I'm currently working on C/++/# programming because I'm planning to attend college for Software Development in hopes of possibly working with some major and minor developers down the line and I was told by a mentor to come here. I'm currently working on a program that uses numbers divisible by 9 here's my code so far I can get it to go up to 8 numbers but one minor problem what if someone wants to just use 4-3 digits? is there anyway of possibly changing this around to suit that purpose without avoiding lowering numbers?

Code:

#include <stdio.h>


#include <stdlib.h>
 
/* Determines whether or not the users integer is divisible by 9 and displays all numbers withing integer*/
 
int main(void)
{
 
    int  num;  /* Users inputed integer */
    int d8; /* 8th digit of the number */
    int d7; /* 7th digit of the number */
    int d6; /* 6th digit of the number */
    int d5; /* 5th digit of the number */
    int d4; /* 4th digit of the number */
    int d3; /* 3rd digit of the number */
    int d2; /* 2nd digit of the number */
    int d1; /* 1st digit of the number */
     
    printf("please enter any digis to see if its divisable by 9. \n");
    scanf("%d", &num);
     
    if ( num % 9 >= 0)
       {
 
               /* Peels off the integer's digits and assigns them to a variable*/
             d8 = num % 10;
             d7 = (num /100) % 10;
             d6 = (num / 1000) % 10;
             d5 = (num / 10000) % 10;
             d4 = (num / 100000) % 10;
             d3 = (num / 1000000) % 10;
             d2 = (num / 10000000) % 10;
             d1 = (num / 10000000) % 10;
              
             /*prints out the intger's digits starting with the rightmost digit */
             printf("rightmost digit: %d\n", d8);
             printf("%d\n", d7);
             printf("%d\n", d6);
             printf("%d\n", d5);
             printf("%d\n", d4);
             printf("%\n", d3);
             printf("%d\n",d2);
             printf("leftmost digit: %d\n", d1);
              
             /* Tells the user the number is divisible by 9 */
             printf("your number is divisible by 9\n");
      }
   else
      {
                /* Peels off the integer's digits and assigns them to a variable*/
             d8 = num % 10;
             d7 = (num /100) % 10;
             d6 = (num / 1000) % 10;
             d5 = (num / 1000) % 10;
             d4 = (num / 10000) % 10;
             d3 = (num / 10000) % 10;
             d2 = (num / 10000) % 10;
             d1 = (num / 100000) % 10;


             /*prints out the intger's digits starting with the rightmost digit */
             printf("rightmost digit: %d\n", d8);
             printf("%d\n",  d7);
             printf("%d\n",  d6);
             printf("%d\n",  d5);
             printf("%d\n",  d4);
             printf("%\n",  d3);
             printf("%d\n",  d2);
             printf("leftmost digit: %d\n", d1);
              
             /* Tells the user the number isnt divisible by 9 */
             printf("your number is not divisible by 9\n");
      }
       
        system("pause");
    return(0);
}

use an array and an index instead of discrete variables to hold the digits. strip off each rightmost digit in the loop until num == 0

Code:

int d[8];
int index;
...

while(num is nonzero) {
   d[iindex] = rightmost digit of current value of num
   num = shift num right so you have a new rightmost digit
   increment index
}
at this point, 'index' tells you how many digits you have

Code:

if ( num % 9 >= 0)

The % operator is the modulus or remainder operator. If you wanted to know whether a number was divisible by 9, simply check the remainder is 0:

Code:

if (num % 9 == 0) {
    // number is divisible by 9
}
else {
    // number is not divisible by 9
}

However, I suppose that is not the method you wanted to use. Rather you want to use something like the "sum all the digits and if that number is divisible by 9, so is the original number" method. Basically, you repeat that until you are left with a 1-digit number. If that is 9, your original number was divisible by 9. If not, it wasn't.

dmh2000 has a good idea for generally isolating each digit, however, if you are after something more like what I suggested, then I wouldn't bother with the array, or the d1...d8 variables you initially used. I think you could get away with two integers, num and digit_sum.

Hmm thanks for the suggestions I'll let you know how this goes I would've replied sooner but I was busy with some family matters

So I actually got it going well thanks to all the suggestions I've also managed to start working on it in an arrayed version as well

Code:

#include <stdio.h>
#include <stdlib.h>
 
/* Determines whether or not the users integer is divisible by 9 and displays all numbers withing integer*/
 
int main(void)
{
 
    int  num;  /* Users inputed integer */
	int digit_sum; 
	
	
	double answer;
     
    printf("please enter any digis to see if its divisable by 9. \n");
    scanf("%d", &num);
    
	answer =(digit_sum , + num) / 9;


    if ( num % 9 >= 0)
       {
 
               /* Peels off the integer's digits and assigns them to a variable*/
            digit_sum = num % 10;
			
			
              
             /*prints out the intger's digits starting with the rightmost digit */
             printf("rightmost digit: %d\n", digit_sum);
			 
            
              
             /* Tells the user the number is divisible by 9 */
             printf("your number is divisible by 9\n");
      }
   else
      {
             
             
             /* Tells the user the number isnt divisible by 9 */
             printf("your number is not divisible by 9\n");
      }
	answer = (digit_sum, +  num) / 9; 
       
        system("pause");
    return(0);
}

Code:

answer =(digit_sum , + num) / 9;

What are you trying to achieve with this line?

Code:

if ( num % 9 >= 0)

This expression will always be true. % 9 returns the remainder of the division by 9 which is a number between 0 and 8.

Bye, Andreas

Originally Posted by AndiPersti

Code:

answer =(digit_sum , + num) / 9;

What are you trying to achieve with this line?

Code:

if ( num % 9 >= 0)

This expression will always be true. % 9 returns the remainder of the division by 9 which is a number between 0 and 8.

Bye, Andreas

You're losing me on this I'm trying to work over a program where a user can enter a number(s) to see if the number/digits are divisible by 9
So going by your post I'm doing something wrong? I haven't ran any additional numbers yet to see if the program was wrong due to some family issues and internet issues but I'm open to any advice from you Andreas or numbers to test out on the program

Code:

#include<stdio.h>#include<stdlib.h>
int main(){
  int num,sum=0, pause ,r;
  printf("Enter a number: ");
  scanf("%d",&num);
  while(num){
      r=num%10;
      num=num/10;
      sum=sum+r;
  }
  printf("Sum of digits of number:  %d",sum);


  system("pause");


  return 0;
}

Okay so this is what I managed to work on with the program as far as adding up the sum of digits together .

I don't know if this was exactly what you were talking about but I didn't feel right just coming back saying I didn't understand without at least having something done and working

That's definitely what I was getting at. You have the logic down, I was just looking for you to put it into a function. Also, you can shorten up your function and get rid of the temp variable r:

Code:

int sum_digits(int num) {
    int sum = 0;

    while (num) {
        sum += num % 10;  // add the right-most digit to sum
        num /= 10;  // divide num by 10, eliminating the right-most digit
    }

    return sum;
}

Now you can use that function in your main logic. There are some hints in post #3 and my EDIT to post #8 for how to do this.

Originally Posted by anduril462

That's definitely what I was getting at. You have the logic down, I was just looking for you to put it into a function. Also, you can shorten up your function and get rid of the temp variable r:

Code:

int sum_digits(int num) {
    int sum = 0;

    while (num) {
        sum += num % 10;  // add the right-most digit to sum
        num /= 10;  // divide num by 10, eliminating the right-most digit
    }

    return sum;
}

Now you can use that function in your main logic. There are some hints in post #3 and my EDIT to post #8 for how to do this.

So wouldn't I need a Do function to start the while loop? That's where I'm lost I can do if, else but yeah

Originally Posted by Benevolence

So wouldn't I need a Do function to start the while loop? That's where I'm lost I can do if, else but yeah

I don't know what a "Do function" is. Do you mean a do-while loop? Like this:

Code:

do {
    call sum_digits
    do some other stuff
} while (some condition is true);

If that's what you're asking about, you do not need a do-while loop. A do-while loop always executes at least once. But if the original number is already 1 digit, no need to do anything. Just use a regular while loop, which may not execute if the initial condition is already true, i.e. if the number is already one digit:

Code:

while (number has more than one digit) {
    sum_digits
    maybe some other stuff
}

cool