How to return True if only one variable of three is True in BASH?

Question

I have three boolean variables:

a|b|c|
1 1 1 False
1 1 0 False
1 0 1 False
1 0 0 True
0 1 1 False
0 1 0 True
0 0 1 True
0 0 0 False

I need a logical expression that returt true only if one variable is true. I try with

if  [[ ( $a == 1 || $b == 1 || $c == 1 ) && ( $a == 1  &&  $b == 1  &&  $c == 1 ) ]];  then
  return True
fi

Thank you

Some language-agnostic solutions are shown here: https://stackoverflow.com/q/3466452/2088135 — Tom Fenech, Sep 19 '19 at 09:52

choroba · Accepted Answer · 2019-09-19T13:45:05.267

7

You can just sum the values and check whether the result equals 1.

Here, I'm using an arithmetic expression to sum the numbers and check the equality.

#! /bin/bash
for a in 0 1 ; do
    for b in 0 1 ; do
        for c in 0 1 ; do
            printf '%s ' $a $b $c
            (( 1 == a + b + c )) && echo True || echo False
        done
    done
done

edited Sep 19 '19 at 13:45

answered Sep 19 '19 at 09:53

choroba

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score 1 · Answer 2 · answered Feb 25 '20 at 09:23

This is the logic which i collected (a^b^c) & ~(a&b&c)

xor all values and remove the case where all values of abc are true truth table

a b c f = a^b^c g = ~(a&b&c) h = f&g

0 0 0 0 1 0

0 0 1 1 1 1

0 1 0 1 1 1

0 1 1 0 1 0

1 0 0 1 1 1

1 0 1 0 1 0

1 1 0 0 1 0

1 1 1 1 0 0

here f is the xor of a b and c and g is the negation of a&b&c the output produces only when one value is true

Have a nice day

How to return True if only one variable of three is True in BASH?

2 Answers2