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Full Discussion: Analysis in bitwise XOR
Top Forums UNIX for Beginners Questions & Answers Answers to Frequently Asked Questions FAQ Submission Queue Analysis in bitwise XOR Post 302608273 by Corona688 on Friday 16th of March 2012 06:11:07 PM
Old 03-16-2012
You are getting a number of things mixed up together and confused.

Quote:
In general, if the operands are in the form of n ^ (n+1) and also n is an even number, then the result is always 1.
xor can be considered a form of comparison, too. What bit is going to change when you add 1 to an even number? The last bit, of course, the 1 bit. This isn't so mysterious or strange.

Your perl examples don't work because they contain syntax errors, logic errors, and you typed them in MS Word, which destroys all code by inserting smart quotes. But perl does, to my surprise, do what you say:

Code:
perl -e "print 'asdf' ^ 'asdf';" | hexdump -C
00000000  00 00 00 00                                       |....|
00000004

 

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integer(3pm)						 Perl Programmers Reference Guide					      integer(3pm)

NAME
integer - Perl pragma to use integer arithmetic instead of floating point SYNOPSIS
use integer; $x = 10/3; # $x is now 3, not 3.33333333333333333 DESCRIPTION
This tells the compiler to use integer operations from here to the end of the enclosing BLOCK. On many machines, this doesn't matter a great deal for most computations, but on those without floating point hardware, it can make a big difference in performance. Note that this only affects how most of the arithmetic and relational operators handle their operands and results, and not how all numbers everywhere are treated. Specifically, "use integer;" has the effect that before computing the results of the arithmetic operators (+, -, *, /, %, +=, -=, *=, /=, %=, and unary minus), the comparison operators (<, <=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, ^, <<, >>, |=, &=, ^=, <<=, >>=), the operands have their fractional portions truncated (or floored), and the result will have its fractional portion truncated as well. In addition, the range of operands and results is restricted to that of familiar two's complement integers, i.e., -(2**31) .. (2**31-1) on 32-bit architectures, and -(2**63) .. (2**63-1) on 64-bit architectures. For example, this code use integer; $x = 5.8; $y = 2.5; $z = 2.7; $a = 2**31 - 1; # Largest positive integer on 32-bit machines $, = ", "; print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y, $y == $z, $a, $a + 1; will print: 5.8, -5, 7, 3, 2, 10, 1, 2147483647, -2147483648 Note that $x is still printed as having its true non-integer value of 5.8 since it wasn't operated on. And note too the wrap-around from the largest positive integer to the largest negative one. Also, arguments passed to functions and the values returned by them are not affected by "use integer;". E.g., srand(1.5); $, = ", "; print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10); will give the same result with or without "use integer;" The power operator "**" is also not affected, so that 2 ** .5 is always the square root of 2. Now, it so happens that the pre- and post- increment and decrement operators, ++ and --, are not affected by "use integer;" either. Some may rightly consider this to be a bug -- but at least it's a long-standing one. Finally, "use integer;" also has an additional affect on the bitwise operators. Normally, the operands and results are treated as unsigned integers, but with "use integer;" the operands and results are signed. This means, among other things, that ~0 is -1, and -2 & -5 is -6. Internally, native integer arithmetic (as provided by your C compiler) is used. This means that Perl's own semantics for arithmetic operations may not be preserved. One common source of trouble is the modulus of negative numbers, which Perl does one way, but your hardware may do another. % perl -le 'print (4 % -3)' -2 % perl -Minteger -le 'print (4 % -3)' 1 See "Pragmatic Modules" in perlmodlib, "Integer Arithmetic" in perlop perl v5.12.1 2010-04-26 integer(3pm)
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