It has nothing to do with floats being compared to integers. In fact, you're comparing floats with floats. 

>perl -e"use Devel::Peek; Dump(3.0)"
SV = NV(0x184968c) at 0x225318      <-- NV = float
  REFCNT = 1
  FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK)
  NV = 3

[download]

Also, it has nothing to do with Perl or Windows. Were you to write the same thing in C or BASIC, you'd get the same results. 

#include <stdio.h>

int main() {
   double x;

   for (x=1.0; x <= 3.0; x += 0.2) {
      printf("%lf\n", x);
   }

   printf("Stop: x=%lf\n", x);
   if (x != 3.0) {
      printf("No, x is not 3.\n");
   }

   return 0;
}

[download]
1.000000
1.200000
1.400000
1.600000
1.800000
2.000000
2.200000
2.400000
2.600000
2.800000
Stop: x=3.000000
No, x is not 3.

[download]

2/10th is also a periodic number in binary. It would take an infinite amount of storage to store it and an infinite amount of time to process it. There is some precision loss.

If you were to print the entire number instead of rounding it (which numerical stringification does), you'd find that you're comparing 3.0000000000000004 to 3.0.

Update: Added C code.

In reply to Re^2: When is a 2 not a 2? by ikegami
in thread When is a 2 not a 2? by RockM66

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