Forums Bug Reports $not

Why $not doesn't return the binary?
ie:
//echo -a $not(1) return: 4294967294
and not 0

//echo -a $not(100) return: 4294967195
and not 011

I need it to make C1 and C2 to make subs in binary like add.

alias binnot return $right($base($not($base($1,2,10)),10,2),$len($1))

This is not a bug. $not() (as well as $or(), $and() etc) accepts decimal values, not binary. You need to use $base() to convert 110 to decimal and then use it as input to $not(). Also note that a number that's to be NOT'd is considered 32-bit. For example, 5 in binary is:
00000000000000000000000000000101
$not(5) is 4294967290, which, in binary, is:
11111111111111111111111111111010
by comparing the above two binary numbers, it's easy to conclude that $not() works as expected.

If I understood what you want to do, it's easy to make an alias for this:

Edit: looks like I was a bit late. While I was previewing my post, theRat sneaked in with a very clever solution that does the same thing.

since $not returns a NOT in a 32bit so when u are doing $not(Num) it first convert the Num to binary and then make it 32bit
for example
$not(2)
binary of 2 is 10 ( $base(2,10,2) )
then make it 32 bit (32 digits)
it's 00000000000000000000000000000010 ( $base(10,10,10,32) )
then compare NOT in each digit (0 NOT = 1, 1 NOT = 0)
it will be
11111111111111111111111111111101
the decimal of this number is 4294967293 ( $base(11111111111111111111111111111101,2,10) )
that's how $not works..

I understand, thankz