Example Of Code

int number = 12345;
int sum = 15;

How do I print the below sentence taking into account that the number can be dynamically chosen by the user using a scanf function and I already have a dynamic function that calculates the sum of each digit.

"1   2   3   4   5 = 15"

Edit: Thank you guys so much for your help!

CodePudding user response：

If the input is coming in as a string, the easiest thing to do is to leave it as a string. Don't ever convert it to an int. (eg, if you are reading the data with scanf using %d, just use %s instead). If the data is already an integer, the easiest thing to do is (probably) to convert it to a string:

#include <stdio.h>
#include <stdint.h>

int
main(void)
{
    int number = 12345;
    char buf[128];
    int sum = 0;
    snprintf(buf, sizeof buf, "%d", number);
    for( char *s = buf; *s; s   ){
        if( s != buf ){
            fputs("   ", stdout);
        }
        putchar(*s);
        sum  = *s - '0';
    }
    printf(" = %d\n", sum);
}

CodePudding user response：

Easiest by far (and probably fastest too actually) is to stick with strings. Here's an example, error handling yet to be implemented:

#include <stdio.h>

int main (void)
{
  char str[128];
  scanf("%s", str);

  int sum = 0;
  for(int i=0; str[i]!='\0'; i  )
  {
    sum  = str[i] - '0'; // convert to integer digit, then add to sum

    printf("%c ", str[i]);
    if( str[i 1] == '\0' )
      printf("= ");
    else
      printf("  ");
  }

  printf("%d", sum);
}

CodePudding user response：

You can try something like this.

#include <stdio.h>  
#define MAX 100

void printIndiviualDigits(int number)  
{  
    int arr[MAX];  
    int i = 0;  
    int j, r,sum=0;

// Till number becomes 0
while (number != 0) {

    // Extract the last digit of number
    r = number % 10;

    // Put the digit in arr[]
    arr[i] = r;
    i  ;

    // Update number to number/10 to extract
    // next last digit
    number = number/ 10;
}

// Print the digit of number by traversing
// arr[] reverse
for (j = i - 1; j > -1; j--) {
    printf("%d", arr[j]);
    sum  = arr[j];
    if(j > 0){
      printf(" ");
    }
}

   printf("=%d",sum);
}

// main function

int main()  
{  
    int number;
    printf("Enter any number:\n");
    scanf("%d",&number);
    printIndiviualDigits(number);
    return 0;
}

CodePudding user response：

This is not the most efficient but it's a charmingly recursive solution.

The challenge isn't totally digits. The challenge is outputting them in decreasing order of significance unless you're always provided with the number as a character string!

Expected Output:

15 -> 1   5 = 6
12345 -> 1   2   3   4   5  = 15
123456 -> 1   2   3   4   5   6 = 21
49 -> 4   9 = 13
-78 -> 7   8 = 15
9 -> 9 = 9
-9 -> 9 = 9
10 -> 1   0 = 1
-10 -> 1   0 = 1

#include <iostream>

int sum(int n){
    if(n==0){
        return 0;
    }
    int r=sum(n/10);
    int d=n;
    if(d<0){
        d=-d;
    }
    r =d;
    if(n>=10 || n<=-10){
        printf("   ");
    }
    printf("%d",d);
    return r;
}

void show(int n){
    printf("%d -> ",n);
    int r=sum(n);
    printf(" = %d\n",r);
}

int main() {
    
    show(15);
    show(12345);
    show(123456);
    show(49);
    show(-78);
    show(9);
    show(-9);
    show(10);
    show(-10);

    return 0;
}


int main() {
    
    show(15);
    show(123456);
    show(49);
    show(-78);
    show(9);
    show(-9);
    show(10);
    show(-10);

    return 0;
}

CodePudding user response：

Simply recursion can be used to print the first digits when the value is 10 or more, along with the " ".

#include <limits.h>
#include <stdlib.h>
#include <stdio.h>

// Print each digit with a trailing "   "
static int sum_helper(int x) {
  int sum = 0;
  if (x >= 10) {
    sum  = sum_helper(x / 10);
    x %= 10;
  }
  printf("%d   ", x);
  return sum   x;
}

void sum_line(int x) {
  printf("d: ", x);
  int last_digit = abs(x % 10);
  int sum = last_digit;
  int first_digits = abs(x / 10);
  if (first_digits) {
    sum  = sum_helper(first_digits);
  }
  printf("%d = %d\n", last_digit, sum);
}

Test code

int main(int argc, char **argv) {
  sum_line(123);
  sum_line(12345);
  sum_line(0);
  sum_line(1);
  sum_line(-12345);
  sum_line(INT_MAX);
  sum_line(INT_MIN);
  return 0;
}

Output

        123: 1   2   3 = 6
      12345: 1   2   3   4   5 = 15
          0: 0 = 0
          1: 1 = 1
     -12345: 1   2   3   4   5 = 15
 2147483647: 2   1   4   7   4   8   3   6   4   7 = 46
-2147483648: 2   1   4   7   4   8   3   6   4   8 = 47