From thj@hpesthj.fc.hp.com

#!/usr/local/bin/perl5 -w

  $x = 22818.875 - 22765.65;
  print("$x\n");

outputs​:

53.2249999999985

The equiv C code

#include <stdio.h>

void main(void)
{
  double x;
  x = 22818.875 - 22765.65;
  printf("%lf\n", x);
}

correctly outputs​:

53.225

From @tamias

On Mon, Jan 10, 2000 at 01​:03​:01PM -0700, Tim Jackson wrote​:

On HP PA-RISC systems (perl version 5.005_02 built for PA-RISC1.1), the
following code

#!/usr/local/bin/perl5 -w

$x = 22818.875 - 22765.65;
print("$x\n");

outputs​:

53.2249999999985

The equiv C code

#include <stdio.h>

void main(void)
{
double x;
x = 22818.875 - 22765.65;
printf("%lf\n", x);
}

correctly outputs​:

53.225

Those two code samples are not equivalent. Try this instead​:

  #!/usr/local/bin/perl5 -w

  $x = 22818.875 - 22765.65;
  printf "%lf\n", $x;

Also, refer to the first entry in perlfaq4.

Ronald