Subtracting columns based on greater than in bash - unix

Question

I have a file with multiple columns:
$ cat file.txt
a bc 34 67
t gh 68 -34
f jh -9 76
h in -66 -14
and so on
I am trying to extract when both columns are negative; when both are positive then subtract the two columns based on which value is greater; and if either column negative then add both the columns
For both negative its quite easy:
less file.txt | egrep -i "\-.*\-" | less
Expected Output:
h in -66 -14
For both positive I tried the following to no avail:
less file.txt | egrep -iv "\-.*\-" | awk '($3>$4 {print $0,($3-$4)}) || ($4>$3 {print $0,($4-$3)})' | less
Expected Output:
a bc 34 67 33
For either negative,
less file.txt | egrep -iv "\-.*\-" | awk '($3<0||$4<0) {print $0,($3+$4)}' | less
Expected Output:
t gh 68 -34 34
f jh -9 76 67
I am seeing this error:
awk: cmd. line:1: (FILENAME=- FNR=3208) fatal: print to "standard output" failed (Broken pipe)
egrep: write error
I know its a basic thing to do, any help would be appreciated!

Answer 1

One awk idea:
awk '
$3<0 && $4<0 { print $0 ; next }
$3>0 && $4>0 { print $0, ($3>=$4 ? $3-$4 : $4-$3); next }
{ print $0, $3+$4 }
' file.txt
NOTE: may need to be tweaked if $3==0 and/or $4==0 ... depends on OP's requirements for this scenario
This generates:
a bc 34 67 33
t gh 68 -34 34
f jh -9 76 67
h in -66 -14

Answer 2

Another awk implementation:
awk '
function abs(x) {
if (x < 0) x = -x
return x
}
$3 >= 0 || $4 >= 0 {$(NF+1) = abs(abs($3) - abs($4))}
{print}
' file.txt
a bc 34 67 33
t gh 68 -34 34
f jh -9 76 67
h in -66 -14
If you wanted to do this in plain bash:
abs() { echo $(($1 < 0 ? -($1) : $1)); }
while read -ra fields; do
a=${fields[2]}
b=${fields[3]}
if ((a >= 0 || b >= 0)); then
fields+=($(abs $(($(abs $a) - $(abs $b)))))
fi
echo "${fields[*]}"
done < file.txt