Why do I get bash integer expression error from [ ${guess[0]} -eq num ]? -

i trying prompt user guess value between 0-9. getting error integer expression expected. don't understand why can explain going wrong?

num=8  echo "enter value between 0-9" read -a guess  if [ ${guess[0]} -eq num ] ;     echo "you guesses right number" elif [ ${guess[0]} -gt num ] ;      echo "you guessed high" elif [ ${guess[0]} -lt num ] ;     echo "you guess low" fi 

to make minimum number of changes necessary code run correctly:

#!/bin/sh  num=8  echo "enter value between 0-9" read guess  if [ "$guess" -eq "$num" ] ;     echo "you guesses right number" elif [ "$guess" -gt "$num" ] ;      echo "you guessed high" elif [ "$guess" -lt "$num" ] ;     echo "you guess low" fi 

notes:

• since you're reading single value, no compelling reason use array -- easy take out -a argument read; doing makes code compatible posix sh, , #!/bin/sh on systems posix-compliant shells.
• you need expand variables when using test (or [). $num, not bare $num; latter possible in numeric context, whereas test regular command as parser concerned -- doesn't create special context @ syntax level.
• because [ regular command, need spaces after it, other command name.
• because [ regular command, need spaces before ] expects last argument -- same when passing arguments else.
• if shell bash (your script started #!/bin/bash), use math context: if (( guess == num )), or if (( guess > num )). is math context, , has special parsing rules.