So sometime human being confuse themselves. As sauoq pointed out, ~~ is nothing more than ~ and ~ again.

However I disagree with one point bart mentioned in his post, in which he said that ~ works differently agasinst integer and string. In fact, it works in exactly the same way against integer and string. This becomes crystal clear if you understand this at the bit level (0=>1, 1=>0), not the human interpretation of those bits: 

use strict;
use warnings;

#integer 32 bit
{
    my $a = 1;
    print "a = $a\n";
    #a: 00000001, ~ - fffffffe
    print 0xfffffffe, "\n";
    print "~a = ", ~$a, "\n";
}
#string
{
    my $a = "abcd";
    (my $a_ord = $a) =~ s/(.)/sprintf "%02x", ord $1/ge;
    print "a = $a\n";
    print "a_ord = $a_ord\n";
    #a:  97, hex - 61, ~ - 9e
    #b:  98, hex - 62, ~ - 9d
    #c:  99, hex - 63, ~ - 9c
    #d: 100, hex - 64, ~ - 9b
    my $b = ~$a;
    (my $b_ord = $b) =~ s/(.)/sprintf "%02x", ord $1/ge;
    print "b_ord = $b_ord\n";
}

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In reply to Re: How does ~~ force string interpolation of code into scalar context? by pg
in thread How does ~~ force string interpolation of code into scalar context? by davido

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