Get Integer input and output it line by line without an array

Question

I am trying to write a c program when I input an integer using scanf and split that integer number and print it line by line without an array. I can show you example how I am going to do that.

123 / 100 = 1
123 % 100 = 23

23 / 10 = 2
23 % 19 = 3

1
2
3

I know how to do that but the problem is when I run this code .

# include <stdio.h>
# include <string.h>
# include <math.h>

int main (void)
{
    int no, a;
    int count, new;
    int newNum = 0;

    printf("Enter an intger number = ");
    scanf("%d", &no);

    newNum = no;
    printf("You entered = %d\n", newNum);

    while(newNum != 0){
        newNum = newNum / 10;
        count++;
    }

    count--;
    count = pow(10, count);

    printf("Power of ten = %d\n", count);

    while(count != 1){
        new = no / count;
        no = no % count;
        printf("%d\n", new);
        count = count / 10;
    }
    return 0;
}

Output:

Enter an intger number = 123
You entered = 123
Power of ten = -2147483648
0
0
0
0
0
0
0
0
-5
-9
Floating point exception (core dumped)

Problem is power of ten line did not output right value But If I comment second while loop section.

# include <stdio.h>
# include <string.h>
# include <math.h>

int main (void)
{
    int no;
    int count, new;
    int newNum = 0;

    printf("Enter an intger number = ");
    scanf("%d", &no);

    newNum = no;
    printf("You entered = %d\n", newNum);

    while(newNum != 0){
        newNum = newNum / 10;
        count++;
    }

    count--;
    count = pow(10, count);

    printf("Power of ten = %d\n", count);

//  while(count != 1){
//      new = no / count;
//      no = no % count;
//      printf("%d\n", new);
//      count = count / 10;
//  }
    return 0;
}

Output:

Enter an intger number = 123
You entered = 123
Power of ten = 100

This time Power of ten shows the correct value.

What can I do to avoid this issue?

and

• Is there any way to do this without using arrays?

Answer 1

In my program I have set my integer variables equal to 0 like this.

int no, a = 0;
int count = 0, new = 0;
int newNum = 0;

After that my program works perfectly, but didn't get the full output. That problem came from this section

while(count != 1){
    new = no / count;
    no = no % count;
    printf("%d\n", new);
    count = count / 10;
}

Output -:

Enter an intger number = 123
You entered = 123
Power of ten = 100
1
2

You can see 3 didn't in output section. There for I have add extra line outside of the while loop to get my expected output. I have add a following.

while(count != 1){
    new = no / count;
    no = no % count;
    printf("%d\n", new);
    count = count / 10;
}
printf("%d\n", no);

Output -:

Enter an intger number = 123
You entered = 123
Power of ten = 100
1
2
3

Now it is completed.

Answer 2

For starters you should be caution and not to make an overflow calculating the power of 10.

Always test your program for boundary values as for example INT_MAX or UINT_MAX.

There is no need to use mathematical functions.

Here you are.

#include <stdio.h>

int main( void ) 
{
    while ( 1 )
    {
        const unsigned int Base = 10;

        printf( "Enter a non-negative number (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        unsigned int divisor = 1; 

        for ( unsigned int tmp = n; !( tmp < Base ); tmp /= Base )
        {
            divisor *= Base;;
        }

        printf( "%u\n", n / divisor );

        for ( ; divisor != 1; divisor /= Base )
        {
            printf( "%u\n", n % divisor / ( divisor / Base ) );
        }

        putchar( '\n' );
    }

    return 0;
}

The program output might look like

Enter a non-negative number (0 - exit): 1
1

Enter a non-negative number (0 - exit): 10
1
0

Enter a non-negative number (0 - exit): 123456789
1
2
3
4
5
6
7
8
9

Enter a non-negative number (0 - exit): 4294967295
4
2
9
4
9
6
7
2
9
5

Enter a non-negative number (0 - exit): 0

Another approach is to use a recursive function as shown here

#include <stdio.h>

void output_digits( unsigned int n )
{
    const unsigned int Base = 10;

    unsigned int digit = n % Base;

    if ( ( n /= Base ) != 0 ) output_digits( n );

    printf( "%u\n", digit );
}

int main( void ) 
{
    while ( 1 )
    {
        printf( "Enter a non-negative number (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        output_digits( n );

        putchar( '\n' );
    }

    return 0;
}

Answer 3

You want to print each digit of a base10 number. The number of digit can be retrieved with this simple formula

int n_digit = (int)log10(x) +1;

To test it: (int)log10(99) +1 = (int)1.9956 +1 = 2 and (int)log10(100) +1 = (int)2 +1 = 3

So you want to print the quotients of each power of 10:

for(int i=n_digit-1; i>=0; i--)
{
    // Calc 10^i without pow function
    int num_10_pow_i = 1;
    for(int j=0; j<i; j++)
        num_10_pow_i *=10;

    printf("N[%d]=",i);
    printf("%d\n", (int)(x/num_10_pow_i));
    x = x-((int)(x/num_10_pow_i) * num_10_pow_i);
}

Please don't use the pow function working with integers.

Read here for more info: Does pow() work for int data type in C?

Answer 4

In your code variable count has local scope (automatic storage duration).

Local scope variables are not initialized due to their stack allocation.

int count=0;

(C99 standard) section 6.7.8 clause 10:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

If an object that has static storage duration is not initialized explicitly, then:

• if it has pointer type, it is initialized to a null pointer;
• if it has arithmetic type, it is initialized to (positive or unsigned) zero;
• if it is an aggregate, every member is initialized (recursively) according to these rules;
• if it is a union, the first named member is initialized (recursively) according to these rules.

Note: you should avoid to use new as a name of a variable.