Question

I am learning C and I have the following code:

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

int main(int argc, char *argv[])
{
  double x;
  printf("x = ");
  scanf("%ld", &x);
  printf("x = %lf\n", x);

  system("PAUSE");  
  return 0;
}

(I am using Dev C4.9, Windows XP SP3)

When I run the above program and entered 5.3; the program printed x = 0.000000

Can anyone explain why is that, please?

Thanks a lot.

Answer 1

The %ld format string means that it’s expecting to read in a long signed int, but you’re passing it instead a double. You should instead use the %lf format specifier to say that you want a double.

Note that for scanf, the l is required for doubles (and is required to be absent for floats), whereas for printf, the l in %lf has no effect: both %f and %lf have the same output for both floats and doubles, due to default argument promotion.