Digit output in C - c

Question

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;
}

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

#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