#!/usr/bin/perl -l

#Program to calculate the range of unsigned integer numbers
#produced from 1 to 32 bits.

use warnings;
use strict;

for(0..32){
 my $bin='0b'.'1'x$_;
 print "$_ : 0 to ",oct $bin;
}


##</code><code>##

#include <stdio.h>
#include <stdlib.h>
main ()
   {
   long int i=1,j=i;

					/* title			*/
   puts("Bits\tRange");
   puts("----\t-----");
					/* O/P data and calc the next
					 * set of values		*/
   for (i=1; i<=32; i++)
      {
      printf("%2ld\t0-%12lu\n", i, j*=2);
      }
   }

##</code><code>##

for(31..64){

##</code><code>##

Binary number > 0b11111111111111111111111111111111 non-portable at ./bits line
        11

##</code><code>##

Integer overflow in binary number at ./bits line 12 (#1)
    (W overflow) The hexadecimal, octal or binary number you have specified
    either as a literal or as an argument to hex() or oct() is too big for
    your architecture, and has been converted to a floating point number.
    On a 32-bit architecture the largest hexadecimal, octal or binary number
    representable without overflow is 0xFFFFFFFF, 037777777777, or
    0b11111111111111111111111111111111 respectively.  Note that Perl
    transparently promotes all numbers to a floating point representation
    internally--subject to loss of precision errors in subsequent
    operations.

##</code><code>##

Binary number > 0b11111111111111111111111111111111 non-portable at ./bits line
        12 (#2)
    (W portable) The binary number you specified is larger than 2**32-1
    (4294967295) and therefore non-portable between systems.  See
    perlport for more on portability concerns.