Boolean operation: (A & B) vs (A * B)

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hi

In Boolean function evaluations using Perl, what is the difference between these two expressions:

#1 Y = (A & B)
#2 Y = (A * B)
[download]

I cannot seem to find documentation on the #2 form of expression, even though I have seen #2 in Boolean expressions used in Perl. For example, this thread has #2 style: parser for evaluating arbitrarily complex boolean function in a string

Bill Murphy

Comment on Boolean operation: (A & B) vs (A * B) Download Code

Replies are listed 'Best First'.
Re: Boolean operation: (A & B) vs (A * B) by LanX (Saint) on Jan 28, 2016 at 13:43 UTC
In mathematics (i.e. boolean algebra) it is common to denote `` for `and` and `+` for `or` IFF the alphabet is restricted to boolean values 0 and 1. NOTE: 1 + 1 is mapped to 1 there. This works in Perl somehow. (1+1 =2 is still true) But don't* ! It's not recommended ... rather a workaround in languages without boolean operators.² Just sketch the truth table to see how it works. Better use `&&` and `\|\|` instead. > `Y = (A & B)` that's definitely wrong, because `&` is for a bitwise `and` not a logical. Cheers Rolf _{(addicted to the Perl Programming Language and ☆☆☆☆ :) Je suis Charlie!} UPDATE to elaborate more: the standard false value in Perl is not 0 but empty string `''` . While the latter will be evaluated to 0 in arithmetic operations it might cause a confusion. `DB<7> print !!0 DB<8> print !!1 1` [download] footnotes ²) and only for simple cases, b/c you risk a number overflow when multiplying true values. `> perl -E "say ((1+1) * (1+1) * (1+1));" 8` [download]	[reply] [d/l] [select]
Re: Boolean operation: (A & B) vs (A * B) by GotToBTru (Prior) on Jan 28, 2016 at 13:40 UTC
The multiplication operator and the bitwise AND are obviously not the same, but if the only values for A and B are 0 or 1, A * B will produce the same truth table as A && B. Update: replaced & with && per LanX. But God demonstrates His own love toward us, in that while we were yet sinners, Christ died for us. Romans 5:8 (NASB)	[reply]
Re: Boolean operation: (A & B) vs (A * B) by hippo (Archbishop) on Jan 28, 2016 at 13:40 UTC
`A * B` is a numerical multiplication. If the (perl) values of A and B can only be 1 and 0 then this is equivalent to a Boolean AND. `A & B` is a bitwise AND. Again, if 1 and 0 are the only values this is also equivalent to a Boolean AND. Whether or not these expressions are entirely equivalent depends on the restrictions of the values to the binary set {0, 1}.	[reply] [d/l] [select]
Re^2: Boolean operation: (A & B) vs (A * B) by LanX (Saint) on Jan 28, 2016 at 13:46 UTC
> If the (perl) values of A and B can only be 1 and 0 then this is equivalent to a Boolean AND. well it's a bit more complicated... 1+1 = 2, see my post Re: Boolean operation: (A & B) vs (A * B) Update as pointed out neither the OP nor hippo used +, I saw the wider scope ... Cheers Rolf _{(addicted to the Perl Programming Language and ☆☆☆☆ :) Je suis Charlie!}	[reply]
Re: Boolean operation: (A & B) vs (A * B) by kennethk (Abbot) on Jan 28, 2016 at 15:58 UTC
All Perl operators are documented in perlop. Relevant for this conversation: `*` -- Multiplicative Operators `&` -- Bitwise And `&&` -- C style Logical And `and` -- Logical And Also, to get parser for evaluating arbitrarily complex boolean function in a string, write `[id://399322]` as documented in Markup in the Monastery. #11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.	[reply] [d/l] [select]
Re: Boolean operation: (A & B) vs (A * B) by u65 (Chaplain) on Jan 28, 2016 at 13:43 UTC
To ease analysis that node link is 399322.	[reply]

Back to Seekers of Perl Wisdom