How to print a float number without extra zero digits in C

Question

When I print a float number with "%f", it prints a float with 6 digits, but I don't want to print extra zeros. I can to use for example "%.3f" to print just 3 digits, but I want to print a float with number of random digits ,and using "%.{number}f" is not my answer. for example in:

#include <stdio.h>
int main()
{
    float a, b;
    printf("Please enter a number: ");
    scanf("%f", &a);
    printf("Please enter another number: ");
    scanf("%f", &b);
    printf("%f", a / b);
    return 0;
}

a/b can have even decimal digits. Of course, my question also applies to double and long double too.

Can you give an example for what you try to do? What is your expected output? — black, Nov 01 '20 at 20:34
Use `"%g"` (or `"%lg"`) which will only print the needed number of digits. (e.g. for `1.0` it will only output `1`, for `1.25` it will output `1.25`) — David C. Rankin, Nov 01 '20 at 20:42
@Weather Vane why are you recommending `sprintf()` instead of `snprintf()`? — DarkAtom, Nov 01 '20 at 23:56
@Weather Vane The OP is obviously a beginner, and you probably shouldn't recommend such practice to beginners, because sooner or later they are going to make the mistake of using it when not appropriate. — DarkAtom, Nov 02 '20 at 00:08
@DarkAtom. So why are recommending not checking the return value from `scanf()` in your code? — Weather Vane, Nov 02 '20 at 00:10
Because the OP is also not doing it. Checking against buffer overflows is much more important then checking for validity of the input in toy programs. — DarkAtom, Nov 02 '20 at 00:13

score 1 · Accepted Answer · answered Nov 02 '20 at 04:18

There are many ways to remove the extra zeros.

Easiest way is to use %g as the format specifier in the printf statement. Eg. printf("%g",a/b);. This is remove the extra zeros after the decimal.
You can use printf("%.0d%.4g\n", (int)(a/b)/10, (a/b)-((int)(a/b)-(int)(a/b)%10));.
You can use a separate function to remove the extra zeros but it is a tedious process. First transfer all the digits to a string and remove the unwanted zeros from the end.
You can also use %.2f format specifier to reduce the zeros. Eg. printf("%.2f",a/b); - this can be used where you want to print the exact number of digits after the decimal point. %.<the number of digits to be printed after decimal point>f.

1 and 3 are good ideas. 2 is prone to failure: 1) when `a/b` outside `int` range, 2) the rounded FP math rounds unexpectedly as with 0.9999999999. — chux - Reinstate Monica, Nov 02 '20 at 04:34

DarkAtom · Answer 2 · 2020-11-02T00:12:00.220

Here is a piece of code to print n non-zero decimal digits:

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

int main(void)
{
    int n;
    float a, b;
    scanf("%f %f %d", &a, &b, &n);
    float result = a / b;
    printf("%d", (int)result);
    int decimal = (int)trunc(result * pow(10, n)) % (int)pow(10, n);
    if (decimal != 0)
        while (decimal % 10 == 0)
            decimal /= 10;
    printf(".%d\n", decimal);
}

How to print a float number without extra zero digits in C

2 Answers2