Please help, need user to decide how many decimal points to print.

I am trying to have a user input a number, 5 for example, and then prints a number i have with that many decimal points.

example:

user types: 5

prints: 1.10293

(i chose random numbers here, but has 5 decimal points)

I only know how to do this printf("%.5lf", mynumber);

and that would print 5, but always 5.

i want it to print the amount the user puts in.

thank you for any help or advice

There's sprintf that prints whatever you want into a string. Print the number you want, along with the format of course, into a string and pass it as the format string of printf.

Quote:

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Originally Posted by GReaper

There's sprintf that prints whatever you want into a string. Print the number you want, along with the format of course, into a string and pass it as the format string of printf.

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hello thanks, I am sorry but I am very new at this and do not understand.

sprintf - C++ Reference

Use it kind-of like this:

Code:

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char buffer[10];

sprintf(buffer, "%%.%df", digits); // This produces "%.Nf", where N stands for the value held by 'digits'
printf(buffer, mynumber); // This takes the string produced above as a format

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@frezzy

I think you were looking for this:

Code:

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

int main(int argc, char *argv[])
{
  double val = 12.3456789;
  int places = 0;

  if(argc != 2)
  {
      printf("USAGE: %s [decimal place int]\n", argv[0]);
      printf("Example: %s 2 (For two decimal places)\n", argv[0]);
      exit(EXIT_FAILURE);
  }
  else
  {
      places = atoi(argv[1]);
  }

  printf("%d Decimal places: %.*lf\n", places, places, val);

  return 0;
}

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Use the '*' char to fill in the field width of the decimal portion in the format string, input by the user.

Oops, completely forgot about that specifier. Use @rstanley's solution, it's better for what you want to do.