Something wrong with %lf in C

Question

I'm doing homework and I have no idea why the %lf selector isn't working. I have to take a line of characters and determine if they can be a floating point or whole number and then print that number. Here's the code:

#include <stdio.h>

int main() {
    char ch;
    int isNumber = 1, dot = 0, negativeMult = 10;
    double result = 0;

    printf("\nEnter characters: ");
    scanf("%c", &ch);
    while (ch != 10) {
        if ((ch >= '0' && ch <= '9')) {
            if (dot) {
                result = result + (ch - '0') / negativeMult;
                negativeMult *= 10;
            } else {
                result = result * 10 + (ch - '0');
            }
        } else
        if (ch == '.')
            if (dot)
                isNumber = 0;
            else
                dot = 1;
        else {
            isNumber = 0;
            break;
        }
        scanf("%c", &ch);
    }

    if (isNumber)
        printf("\nThe number is %lf", result);
    else
        printf("\nEntered characters are not able to be a number.");

    return 0;
}

Edit: I forgot output. Sorry. Input: Enter characters: 123.648

Output: The number is 123.000000

`(ch - '0') / negativeMult` is computed as `int` so it will usually come out as zero, try making negativeMult a double. — , Nov 19 '21 at 13:54
Did you check your values in a debugger or what makes you think you have a problem with printing? In most cases the previous calculation contain more errors than the part printing. — Gerhardh, Nov 19 '21 at 13:55
Avoid magic numbers like `10` which you use in `while (ch != 10)`. Make that `while (ch != '\n')` instead. — Ted Lyngmo, Nov 19 '21 at 13:57
@FredLarson yes, I know about that, I just forgot. I'm still gonna give it a read tho. — ProGamer2711, Nov 19 '21 at 13:58
"_they_"? Don't listen to those people. What is the purpose? To obfuscate the code? It doesn't run any faster if that's what "_they_" think. — Ted Lyngmo, Nov 19 '21 at 14:02
Btw, as noted above, a possible fix is to make one of the operands in the division a `double` like `result += (ch - '0') / (double)negativeMult;` — Ted Lyngmo, Nov 19 '21 at 14:03
You should change the expression for decimal numbers to: `result = result * 10 + ch - '0';` in all cases and divide by `negativeMult` only after the last digit. `negativeMult` should be initialized to `1` instead of `10`. — chqrlie, Nov 19 '21 at 14:06
@AlexanderShestakov *I know, they made me* Frankly, "they" are not very bright if they think `10` is better than `'\n'`. — Andrew Henle, Nov 19 '21 at 14:15

score 0 · Accepted Answer · answered Nov 19 '21 at 14:07

the error is here:

result = result + (ch - '0') / negativeMult;

(ch - '0') / negativeMult is integer division and it is always 0

it has to be

result = result + (double)(ch - '0') / negativeMult;

some more small errors amendments:

int main(void)
{
      char ch;
        int isNumber = 1, dot = 0, negativeMult = 10;
        double result = 0;
        int scanfresult;

        printf("\nEnter characters: ");
        scanfresult = scanf("%c", &ch);
        while (ch != '\n' && scanfresult == 1)
        {
                if ((ch >= '0' && ch <= '9'))
                {
                        if (dot)
                        {
                                result = result + (double)(ch - '0') / negativeMult;
                                negativeMult *= 10;
                        }
                        else
                        {
                                result = result * 10 + (ch - '0');
                        }
                }
                else if (ch == '.')
                        if (dot)
                                isNumber = 0;
                        else
                                dot = 1;
                else
                {
                        isNumber = 0;
                        break;
                }
                scanfresult = scanf("%c", &ch);
        }

        if (isNumber)
                printf("\nThe number is %f", result);
        else
                printf("\nEntered characters are not able to be a number.");

        return 0;
}

https://godbolt.org/z/nTKdjYsz8

Something wrong with %lf in C

1 Answers1