I got the expected results below, but I didn't get the expected expansion in case (3).
I expected to see "+ [[ 0 -eq 0 ]]" and not "+ [[ 0 -eq r ]]". With set -x, aren't the left and right parameters in a expression the final result of all expansions, including arithmetic expansion? Isn't that enough to change r to 0 in case (3)? (1) set -xv declare r=0 /bin/true; if [ $? -eq r ]; then echo "1 Y";else echo "N";fi ... + '[' 0 -eq r ']' ./t3: line 12: [: r: integer expression expected + echo N (2) /bin/true; if [ $? -eq $r ]; then echo "2 Y"; else echo "N";fi ... + '[' 0 -eq 0 ']' + echo '2 Y' (3) /bin/true; if [[ $? -eq r ]]; then echo "3 Y"; else echo "N";fi ... + [[ 0 -eq r ]] <----- + echo '3 Y' (4) /bin/true; if [[ $? -eq $r ]]; then echo "4 Y"; else echo "N";fi ... + [[ 0 -eq 0 ]] + echo '4 Y' (Similar to 3) unset r a; a=0; declare -i r=a /bin/true; if [[ $? -eq r ]]; then echo "5 Y"; else echo "N";fi ... + [[ 0 -eq r ]] <----- + echo '2 Y' Thank you. Peggy Russell
