expr bit-wise "and" operator, dual of |

Arguments must be integers, result is an integer.

Bit n of the result is 1 if bit n of each argument is 1. Otherwise, bit n of the result is 0.

For negative arguments, we use the extended definition of ~ that ~$a==-1-$a. There are then the following cases:

Case Result
$a>=0,$b>=0 Ordinary bitwise &
$a>=0,$b<0 $a&$b == $a & ~(~$b) Contrapositive law
$a&$b == $a & ~ (-1-$b) Extended definition of ~
Since -1-$b is positive, $a & ~(-1-$b) can be evaluated in bitwise fashion.
$a<0,$b>=0 Commute to ($b & $a) and use the calculation above
$a<0,$b<0 $a&$b == ~ (~$a | ~$b) De Morgan's Law
$a&$b == ~ ((-1-$a) | (-1-$b)) Extended definition of ~
$a&$b == -1-((-1-$a) | (-1-$b)) Extended definition of ~
Since -1-$a and -1-$b are both positive, the expression ((-1-$a) | (-1-$b)) can be evaluated in the ordinary bitwise fashion.

For logical/short-cut "and" use the && operator.


AMG: Be wary of the precedence of this operator! Like &&, it has lower precedence than the comparison operators. For more information on why, read the section "Neonatal C" found at [L1 ].

Expression ResultComment
expr 5&2==2 1 expression without parentheses
expr 5&(2==2)1 actual behavior of Tcl and C
expr (5&2)==20 naively expected behavior

Convert negative integer to hexadecimal?

jennylv - 2012-08-07 16:15:14

I want to convert signed integer to 32-bit hexadecimal.

The following works fine for positive integer:

format %.8x $the_decimal_value

But for negative integer, I got 64-bit hexadecimal.

For example 99 is converted to 00000063, but -81 is converted to ffffffffffffffaf.

Anyone know how to get 32-bit hexadecimal fro negative integer? Thanks in advance!

AMG: The %.8x means to pad to eight characters only if narrower than eight to begin with. You want to truncate the value. What you're looking for is the & operator. Try this:

% format %.8x [expr {$the_decimal_value & 0xffffffff}]

jennylv - 2012-08-07 16:28:19

Thanks so much! That works fine:)