Digit output in C

user_17361403 0
Viewed 82

Hello everyone I'm a beginner in coding and I try to figure out how to output all the digits. Example in the number 158 the number 1,5,8,15,58,158. Sorry for the bad English.

I have tried something but it doesnt work for all numbers plus i believe there must be a better way to code it without all the while loops.

    #include <stdio.h>

int main(){
    long num = 5025;
    int num1=num ,num2= num, num3=num;
   
    while(num1 !=0)
    {
        int digit = num1 % 10;
        num1 = num1/10;
        printf("%d\n", digit);
    } 
    while(num2 >10)
    {
        
        int digit = num2 % 100;
        
        num2 = num2 / 10;
                
        printf("%.2d\n", digit);
    }
    while(num3 >100)
    {
        
        int digit = num3 % 1000;
        
        num3 = num3 / 10;
                
        printf("%.3d\n", digit);
    }
    
    return 0;
}
c
4 Answers

One could print to a string and then post its various combinations

  long num = 158;
  char buf[25];
  snprintf(buf, sizeof buf, "%ld", num);

  for (int first = 0; buf[first]; first++) {
    for (int last = first; buf[last]; last++) {
      int width = last - first + 1;
      printf("%.*s\n", width, buf + first);
    }
  }

Output

1
15
158
5
58
8

Depending on the value, you may get repeats. OP has not yet defined what to do in that case.

For a math only approach, I'd use recursion, yet I expect that is beyond OP's ken at this time.

user_24103590

Here is function printing it for any unsigned int number without fancy printf formats. It also prints in the order from shortest to longest:

void print(unsigned num)
{
    char nums[20];
    size_t len;

    len = sprintf(nums, "%u", num);

    for(size_t seql = 1; seql <= len; seql++)
    {
        for(size_t ndig = 0; ndig < len - seql + 1; ndig ++)
        {
            for(size_t dig = 0; dig < seql; dig++)
            {
                printf("%c", nums[dig + ndig]);
            }
            printf("%c", seql == len ? '\n' : ',');
        }
    }
}

int main(void)
{
    print(15895678);
}

Or without strings:

char getNthDigit(unsigned num, size_t digit, size_t len)
{
    size_t pos = len - digit;
    while(--pos) num /= 10;
    return num % 10;
}

size_t getNdigits(unsigned num)
{
    size_t count = 0;
    while(num) 
    {
        count++;
        num /= 10;
    }
    return count;
}


void print(unsigned num)
{
    char nums[20];
    size_t len;

    len = num ? getNdigits(num) : 1;

    for(size_t seql = 1; seql <= len; seql++)
    {
        for(size_t ndig = 0; ndig <= len - seql; ndig ++)
        {
            for(size_t dig = 0; dig < seql; dig++)
            {
                printf("%c", '0' + getNthDigit(num,dig + ndig, len));
            }
            printf("%c", seql == len ? '\n' : ',');
        }
    }
}

https://godbolt.org/z/Wa8nKns9E

user_61100940

My take, without strings and a bit less complexity. If the number has a 0 digit, it will print duplicates for the digit to the right.

#include <stdio.h>

int main(void)
{
    int digits[64]; // size of array must cover however many digits are in long max
    int numDigits = 0;
    long num = 15876;

    // populate array with each digit
    while (num)
    {
        digits[numDigits++] = num % 10;
        num /= 10;
    }

    // start at the end of the array (since digits are loaded in it backwards)
    for (int first=numDigits-1; first>=0; first--)
    {
        int temp = digits[first];
        printf("%d\n", temp);  // print the initial condition
        for (int last=first-1; last>=0; last--)
        {
            // continue to build temp by multiplying by 10 and adding the next digit
            temp = (temp * 10) + digits[last];
            // could easily put the output in a comma-separated list if that's needed
            printf("%d\n", temp);
        }
    }
}

Output

1
15
158
1587
15876
5
58
587
5876
8
87
876
7
76
6

Demo

user_34767800 
#include <stdio.h>

int main(void) {
    int num = -158;
    char text[12];
    sprintf(text, "%d", abs(num));
    
    for(int w=0; w<strlen(text);   ++w)
    for(int s=0; s<strlen(text)-w; ++s)
    printf("%.*s\n", w+1, text+s);
    
    return 0;
}

From IDEOne

Output

Success #stdin #stdout 0s 5632KB
1
5
8
15
58
158
user_348240
Related