If statement is not following its conditional

Question

In my code to roll you just write r and then hit enter but it seems not to read that and go to else that restarts the while loop. The only way to get it to roll is by typing som

Answer 1

Okay, a few comments about your code.

• This is really hard to read because of the lack of indenting. When nesting things, indent what is nested. With indents, you can see where your loops start and end, what code belongs to which if/else, etc.
• In bash, you should always quote your variables to avoid accidental expansion. For example, what happens if someone enters an asterisk (*) instead of an "r"? Your if statement will do wondrous and mysterious things.
• You're using operators wrong. In bash, using single square brackets with if, you compare string equivalency with a single equals (=), not double. And if you want numerical equality, you have -eq. Though you might want to look at bash's extended tests, using double square brackets. (Check the man page for details.)
• Try not to use external tools for things that bash can do by itself. bc, for example, is not needed for integer arithmetic.

So ... all said, here's a code segment of yours, re-written a bit.

while [ "$pt" -eq 1 ]; do

    read -p "Roll or stay (r/s)? " choice

    if [ "$choice" = r ]; then

        die=$(($RANDOM%6+1))

    elif [ "$die" -eq 1 ]; then

        p1s=$((p1s - count))
        echo "You rolled a 1. Your score is $p1s"
        echo "$p2 turn now."
        sleep 1 
        count=0 
        pt=2    

    elif [ $die -gt 1 ]; then

        p1s=$((p1s + die))
        count=$((count + die))
        echo "You rolled a $die. Your score is $p1s"
        pt=1    

    else

Note that I'm not making any claims about whether your program logic is sound.

And what the heck is num.sh? Is it important?