Cure
User
Apr 20, 2000, 4:49 PM
Post #6 of 10
(4295 views)
|
|
Re: "bits and bits" !~ /scambled eggs for brain/;
[In reply to]
|
Can't Post
|
|
Hi
You go out of your way to produce five-digit strings for $thiskey, and
$binary is a five-digit string, so you seem to be doing a *stringwise*
AND. But the code is doing a *bitwise* AND, and the result happens --
purely by chance! -- to look like the result of a stringwise AND, except
for the case of '01000'. The clue is that $truth is being printed as an
integer, not as a five-digit string.
When I stringify $thiskey explicitly (usefully double-quoting a simple
variable!):
my $truth = $binary + "$thiskey";
the 104 becomes 00000 (and all the other TRUTHs are five-digit strings,
as expected).
The reason I am baffled is this:
print "Padded key : $thiskey\n";
# In the line above, $thiskey is a five-digit string.
my $truth = $binary & $thiskey;
# In the line above, $binary is a five-digit string.
print "The TRUTH is $truth\t ($binary and $thiskey )\n";
# In the line above, $thiskey is a five-digit string.
Why does $thiskey become an integer, which 'integrifies' the value of
$binary, for the duration of the conjuction statement, but is still a
string after it???
In binary, 00110 is 0001101110
01000 is 1111101000
The bitwise AND is 0001101000
which is 104
Because if you do the same conversions and bitwise arithmetic as I did
above, you will get BY PURE COINCIDENCE a binary result which, when
converted to decimal, looks the same as a string-wise AND.
|
MAIN
SEARCH
WHO'S 
LOG 