Displaying the digits of an integer without using a reverse number or arrays

Question

Printing an integer each digit at a time in english form recursively

How to get the digits of a number without converting it to a string/ char array?

This is for coding in C, not C++. My knowledge of C is very limited, because I'm in the entry level course, and we just got past midterm. Try to keep this as simple as possible, because I can't include keywords or operators that we haven't covered in class. I don't think this is necessary, since I think it's just my logic that needs help, not my code.

Having referenced the above two examples to write my code for a class, I am stumped as how to finish up my last little piece of the puzzle. I have found a couple of questions-answers here on SO that appeared to be relevant, however they used code to solve the problem that I have no knowledge on. Hopefully, someone can help me with my logic here.

My goal of my assignment is to:

Take a user defined integer, and display the digits in english. For example:

Please enter an integer: 123
You have entered: One Two Three

Then, I need to add up the sum of the digits (and if digits<10, display in english). In this case:

The sum of the individual digits is: Six

Finally, I need to then average the digits using 2 decimal places. In this case:

The average is: 2.00

I have ALL of this completed. Except: My first step lists the digits backwards! It reads 10s place, 100s place, 1000s place, etc. For example:

Please enter an integer: 123
You have entered: Three Two One

My conditions for this portion of the assignment, is that I may use only one switch statement, and I have to use a switch statement (meaning the need for a loop (I went with do)). I may also not use arrays. But finally, and most importantly, I may NOT reverse the input number (which was the solution to the the first version of this assignment). If I could do that, I would not be here.

Here is the excerpt of code that is relevant.

#include <stdio.h>

int main(void)

{

    int userinput, digit

    printf("Please input a number:");
    scanf("%d", &userinput);

    printf("You have entered: ");

    if (userinput < 0)
    {
            printf("Negative ");
            userinput = -userinput;
    }

    do
    {
            digit = userinput%10;
            switch (digit)
            {
                    case 0:
                    {
                            printf("Zero ");
                            break;
                    }
                    case 1:
                    {
                            printf("One ");
                            break;
                    }
                    case 2:
                    {
                            printf("Two ");
                            break;
                    }
                    case 3:
                    {
                            printf("Three ");
                            break;
                    }
                    case 4:
                    {
                            printf("Four ");
                            break;
                    }
                    case 5:
                    {
                            printf("Five ");
                            break;
                    }
                    case 6:
                    {
                            printf("Six ");
                            break;
                    }
                    case 7:
                    {
                            printf("Seven ");
                            break;
                    }
                    case 8:
                    {
                            printf("Eight ");
                            break;
                    }
                    case 9:
                    {
                            printf("Nine ");
                            break;
                    }
                    default:
                    {
                            break;
                    }

            }

            userinput = userinput/10;

    } while (userinput > 0);

    printf("\n");

I don't know how to use unsigned, which means I'm not allowed to. And like I said, I'm really entry level. So I don't actually know what your hint means :( — Hamberglar, Nov 09 '13 at 19:17

score 1 · Answer 1 · answered Nov 09 '13 at 17:56

1

If you can't use an array, use recursion:

void print_textual(int n)
{
    if (n > 9) {
        print_textual(n / 10);
    }

    switch (n % 10) {
    case 0: printf("zero "); break;
    case 1: printf("one "); break;
    case 2: printf("two "); break;
    case 3: printf("three "); break;
    case 4: printf("four "); break;
    case 5: printf("five "); break;
    case 6: printf("six "); break;
    case 7: printf("seven "); break;
    case 8: printf("eight "); break;
    case 9: printf("nine "); break;
    }
}

By the way, this would really be much better if you could use an array for at least the digit names:

void print_textual(int n)
{
    if (n > 9) {
        print_textual(n / 10);
    }

    static const char *names[] = {
        "zero",
        "one",
        "two",
        "three",
        "four",
        "five",
        "six",
        "seven",
        "eight",
        "nine"
    };

    printf("%s ", names[n % 10]);
}

answered Nov 09 '13 at 17:56

Might print unexpected strings if n < 0? – chux - Reinstate Monica Nov 09 '13 at 18:59
@chux I don't believe that's legal input based on the example OP provided. – Nov 09 '13 at 19:08
I'm not sure how this works, so could someone explain it? Also, yes. I need < 0 input to be usable. As you can see by the first if statement, I even check for that so I can print "Negative " and change the integer to a positive number. – Hamberglar Nov 09 '13 at 19:19
@Hamberglar Then you can leave that part in your code and it will work for negative numbers as well. It works like this: if you used modulo arithmetic and division in a loop, that would print the numbers in reverse order (starting with the LSD to the MSD). Now if you instead recurse **before** actually printing the digit, it will reverse the order in which they are printed, so you get back the original order. – Nov 09 '13 at 19:24
I had this problem with the comments on my question, too. I don't actually know what recursion is... Edit: To clarify, this means I can't use recurse either. – Hamberglar Nov 09 '13 at 19:25
@Hamberglar If you can't use neither recursion, nor an array, then you can't solve this problem. – Nov 09 '13 at 19:27
Considering this is an assignment given to me for a class, I don't believe that's possible. There must be a way. – Hamberglar Nov 09 '13 at 19:31
@Hamberglar I suspect that the rest of your class knows what either of an array or recursion is... – Nov 09 '13 at 19:33
To clarify, the only requirement for how I complete the assignment is to use a switch statement and only one switch statement, and not to use arrays or anything we haven't covered in class. It doesn't necessarily need to use the logic I used. That's just the only way I could think of to get something worthwhile to print. – Hamberglar Nov 09 '13 at 19:33
I actually do know what arrays are, but it was specified that we can't use them, because we learned them after this assignment was given. If there's a way to use recursion without the recurse keyword, then that's what needs to be done, but I don't know how to do that since we haven't covered recursion. – Hamberglar Nov 09 '13 at 19:34
@Hamberglar You could as well use an infinitely long `switch` statement, enumerating all possible integers. And then you used only a switch statement. But I'm pretty confident that's not a requirement. – Nov 09 '13 at 19:34
@Hamberglar What "recurse keyword"? – Nov 09 '13 at 19:35
I have no idea how to use recursion, so I assumed when you said "recurse" that meant a keyword. When I looked up recursion, that seemed to be confirmed because they used recurse as a keyword. – Hamberglar Nov 09 '13 at 19:37
@Hamberglar No. Recursion means that a function calls itself. There's no such keyword in C, nor is it necessary. – Nov 09 '13 at 19:38
Then I'm going to look up recursion and see if it fits within the parameters of my knowledge. Thank you for the help and sorry for misunderstanding! – Hamberglar Nov 09 '13 at 19:39

chux - Reinstate Monica · Answer 2 · 2013-11-09T21:46:17.537

To avoid using recursion, form a power-of-10 multiplier scaled to n. Then use this multiplier to determine the digits in most significant to least significant order.

Use the same digits_in_english() to print the digit sum.

void digits_in_english(const char *prompt, int n, int *Sum, int *Count) {
  fputs(prompt, stdout);
  *Sum = 0;
  *Count = 0;
  if (n < 0) {
    printf(" Negative");
    n = -n;
  }
  int m = n;
  int pow10 = 1;
  while (m > 9) {
    m /= 10;
    pow10 *= 10;
  }
  do {
    static const char *Edigit[] = { "Zero", "One", "Two", "Three", "Four",
       "Five", "Six", "Seven", "Eight", "Nine" };
    int digit = n / pow10;
    *Sum += digit;
    (*Count)++;

    // OP knows how to put a switch statement here instead of printf()
    printf(" %s", Edigit[digit]);

    n -= digit * pow10;
    pow10 /= 10;
  } while (pow10 > 0);
  fputs("\n", stdout);
}

void Etest(int n) {
  int Count, Sum;

  // Change this to a printf() and scanf()
  printf("Please enter an integer: %d\n", n);

  digits_in_english("You have entered:", n, &Sum, &Count);
  double Average = (double) Sum / Count;

  // Do no care about the resultant Sum, Count
  digits_in_english("The sum of the individual digits is:", Sum, &Sum, &Count);

  printf("The average is: %.2f\n\n", Average);
}

sample

Please enter an integer: -2147483647
You have entered: Negative Two One Four Seven Four Eight Three Six Four Seven
The sum of the individual digits is: Four Six
The average is: 4.60

Does not work for INT_MIN. A bit tricky to do so in a portable way.

It appears OP is not allowed to use arrays. Hope that does not include strings.

Unfortunately, that does include strings :( – Hamberglar Nov 14 '13 at 08:47 — Hamberglar, Nov 14 '13 at 08:47

score -2 · Accepted Answer · answered Nov 14 '13 at 10:46

So, through a lot of stumbling around, I managed to make my code work the way it was intended using only the knowledge I was allowed to use. Here's the finished product (unless someone sees any huge errors):

    //initializing variables, of course
    int userinput, number1, number2, numbersum;
    int div = 1;
    float number3;

    //displaying instructions to the user
    printf("Please input a number: ");
    scanf("%d", &userinput);

    printf("You have entered: ");

    if (userinput < 0)
    {
            printf("Negative ");
            userinput = -userinput;
    }

    //the variables number1-3 are for the data analysis at the end
    //I am preserving the original input in them so I can mutilate it in the following step
    number1 = userinput;
    number2 = userinput;

    while (div <= userinput)
    {
            div = div*10;
    }

    do
    {
            if (userinput != 0)
            {
                    div = div/10;

                    switch (userinput/div)
                    {
                            case 0:
                            {
                                    printf("Zero ");
                                    break;
                            }
                            case 1:
                            {
                                    printf("One ");
                                    break;
                            }
                            case 2:
                            {
                                    printf("Two ");
                                    break;
                            }
                            case 3:
                            {
                                    printf("Three ");
                                    break;
                            }
                            case 4:
                            {
                                    printf("Four ");
                                    break;
                            }
                            case 5:
                            {
                                    printf("Five ");
                                    break;
                            }
                            case 6:
                            {
                                    printf("Six ");
                                    break;
                            }
                            case 7:
                            {
                                    printf("Seven ");
                                    break;
                            }
                            case 8:
                            {
                                    printf("Eight ");
                                    break;
                            }
                            case 9:
                            {
                                    printf("Nine ");
                                    break;
                            }
                            default:
                            {
                                    break;
                            }

                    }

                    userinput = userinput%div;

            }

            else
            {
                    printf("Zero");
            }

    } while (userinput > 0);

    //line break to make it look pretty
    printf("\n");

    //boring math to determine the sum of the digits
    //assuming all are positive due to know contrary instructions
    //set equal to zero since this variable refers to itself in the following function
    numbersum = 0;

    while (number1 > 0)
    {
            numbersum = numbersum + (number1 % 10);
            number1 = number1 / 10;
    }

    //nested switch in if statement to print english if digits less than or equal to 10
    if (numbersum <= 10)
    {
            switch (numbersum)
            {
                    case 0:
                    {
                            printf("The sum of the individual integers is: Zero");
                            break;
                    }
                    case 1:
                    {
                            printf("The sum of the individual integers is: One");
                            break;
                    }
                    case 2:
                    {
                            printf("The sum of the individual integers is: Two");
                            break;
                    }
                    case 3:
                    {
                            printf("The sum of the individual integers is: Three");
                            break;
                    }
                    case 4:
                    {
                            printf("The sum of the individual integers is: Four");
                                    break;
                    }
                    case 5:
                            {
                            printf("The sum of the individual integers is: Five");
                            break;
                    }
                    case 6:
                    {
                            printf("The sum of the individual integers is: Six");
                            break;
                    }
                    case 7:
                    {
                            printf("The sum of the individual integers is: Seven");
                            break;
                    }
                    case 8:
                    {
                            printf("The sum of the individual integers is: Eight");
                            break;
                    }
                    case 9:
                    {
                            printf("The sum of the individual integers is: Nine");
                            break;
                    }
                    case 10:
                    {
                            printf("The sum of the individual integers is: Ten");
                    }
                    default:
                    {
                            break;
                    }

            }

            printf("\n");
    }

    //else if greater than 10, just print the decimal number
    else
    {
            printf("The sum of the individual digits in the integer is: %d\n", numbersum);
    }

    if (numbersum == 0)
    {
            printf("The average is of zero is not a number.\n");
    }

    else
    {

            //initializing a variable here because it's totally irrelevant to the above functions
            //and this feels cleaner because of it. I'm not sure if this is bad etiquette
            int i;

            //picks out the number of digits in the input and effectively sets i to that number
            for (i = 0; number2 > 0; i++)
            {
                    number2 = number2/10;
            }

            //this is necessary for turning number3 into an actual floating point, not an int stored as float
            number3 = numbersum;

            //math to determine average (sum of digits divided by number of digits)
            number3 = number3 / i;

            printf("The average is: %.2f\n", number3);

    }

    return 0;

It's big and it's probably kind of sloppy (due to how new I am), but it works and that's all that really matters to me right now.

Thanks for the help, guys.

1) "if greater than 10, just print the decimal number" is taking the easy way out. Just loop like the previous major loop. 2) `numbersum == 0` is not needed. "0" should be a valid entry as the average digits is 0. 3) _Somehow_ the 2 tasks that need an integer to textual output should use the same code (a function). Replicating code (the 2 big switch statements) is a coding technique of _last_ resort. 4) Programming tasks "do this without using a feature of the language" are value limited. I've had to code without access to typical _functions_, but not w/o core language feature like _arrays_. — chux - Reinstate Monica, Nov 14 '13 at 15:21
Actually, it was specified in the instructions that I can take the "easy way out" so I jumped on that. I think you might be right about the 0 part. I think i was overthinking that (thought a div/0 thing might be possible). — Hamberglar, Nov 14 '13 at 21:30
Right that `i=0` and `number3 / i` is a non-no. It can be avoided with `i=0; do { i++; number2 = number2/10; } while (number2 > 0)`. Remember this idiom to allow 0 to loop once, but still be a terminating condition. I used the `do { ...} while (pow10 > 0)` in my answer for the same reason, as you did in your `do { } while (userinput > 0)`. — chux - Reinstate Monica, Nov 14 '13 at 23:56

Displaying the digits of an integer without using a reverse number or arrays

3 Answers3