How do I output a result to the number of decimal-points user previously entered (C)?

Question

I'm trying to output my results to the decimal accuracy the user has input. In other words, if they enter 145, then it outputs to 0 d.p. - if they enter 145.67, it outputs the result to 2 d.p.

I've tried achieving this by trying different things with %.lf but didn't manage to do it so far. Is there a way to do this in C? If so, is there a name for it - so I can learn more about it.

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

int main() {
  // Using long, double or something else?
  double number, fourthRoot;

  printf("Enter a number: ");
  scanf("%lf", &number);

  fourthRoot = sqrt(sqrt(number));

  printf("4th root of number %lf is %.10lf", number, fourthRoot);

  return 0;
}

There is no way `scanf` is going to work for you. You need to parse the input more carefully. — William Pursell, Oct 15 '21 at 16:19
What you've asked seems perfectly reasonable, and many have asked it before — but the surprising fact is, it's basically fundamentally impossible. If the user types 145.67, the internal `double` value you actually get will be more like 145.6699999999999874944, and once you've got that number, there's no way to know what the user actually typed. — Steve Summit, Oct 15 '21 at 16:23
One technique to know about is `%g`. That will show good precision, it will strip unnecessary trailing 0's, and it will switch to scientific notation, if necessary, for really big or really small numbers. — Steve Summit, Oct 15 '21 at 16:27
One more question: do you want to preserve the same number of *places past the decimal*, or the number of *significant digits*? What should the result be if the user enters 0.0000000000000625? — Steve Summit, Oct 15 '21 at 16:36
The nameless thing you're looking for is "Read a floating-point number from the user *and* tell me how much precision it had as the user typed it". But it doesn't have a name, because it doesn't exist as standard functionality, although chux's and William Pursell's answers show you how you might try to explicitly achieve it. — Steve Summit, Oct 15 '21 at 16:42

score 0 · Answer 1 · answered Oct 15 '21 at 16:33

Read user input as a line with fgets() or some scanf("%s",...-like code.

char buf[400];
if (fgets(buf, sizeof buf, stdin)) {

And then process the string. One way is with sscanf()

  int n;
  // Get the number and offset to end of scan
  if (sscanf(buf, "%lf%n", &number, &n) == 1) {
    // Look for '.'
    char *dp = strchr(buf, '.');
    if (dp && dp < buf + n) {
      // decimal point found
      int pre = buf + n - dp - 1;
      printf("%.*f\n", pre, number);

This works OK but does get fooled if exponential notation used. Additional code could look for the E or e (or P, p with hexadecimal FP notation).

William Pursell · Answer 2 · 2021-10-15T17:25:20.917

There are a lot of edge cases that you should worry about, but this should get you started:

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

static int
compute_root(const char *s)
{
    char *end;
    double number = strtod(s, &end);
    if( *end ){
        fprintf(stderr, "Invalid input: %s\n", s);
        return 1;
    }
    char *dp = strchr(s, '.');
    int precision = dp ? end - dp - 1: 0;
    char fmt[256];
    snprintf(fmt, sizeof fmt, "4th root of %%s is %%.%dlf\n", precision);
    printf(fmt, s, pow(number, .25));
    return 0;
}

int
main(int argc, char **argv)
{
    int errors = 0;
    char buf[64];
    if( argc > 1 ){
        while( *++argv ){
            errors += compute_root(*argv);
        }
    } else while( scanf("%63s", buf) == 1 ){
        errors += compute_root(buf);
    }
    return errors ? EXIT_FAILURE : EXIT_SUCCESS;
}

As an exercise, consider inputs like 5.345e4. The above code will output with a precision of 5, but you probably want 3. Handling those cases is non-trivial (for certain definitions of "trivial").

How do I output a result to the number of decimal-points user previously entered (C)?

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