如何将结果输出到用户先前输入的小数点数 (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 dp - if they enter 145.67, it outputs the result to 2 dp

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;
}

Answer 1

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).

Answer 2

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").