Why does bash parse same condition in different ways? [closed]

Question

Today I run into strange bash behaviour, i made this simple example:

#!/bin/bash

if true or ! true; then
    echo a;
else
    echo b;
fi

if true -o ! true; then
    echo a;
else
    echo b;
fi

if true || ! true; then
    echo a;
else
    echo b;
fi

if true or 
   ! true; then
    echo a;
else
    echo b;
fi

if true -o 
   ! true; then
    echo a;
else
    echo b;
fi

if true || 
   ! true; then
    echo a;
else
    echo b;
fi

The output is something unexpected for me:

a
a
a
b
b
a

I was prompted that cases 4 and 5 are treated like that:

pi@raspberrypi:~ $ if true or; ! true; then echo a; else echo b; fi
b
pi@raspberrypi:~ $ if true -o; ! true; then echo a; else echo b; fi
b

This is working code, but it looks like a syntax error for me, and in the 6th case this is it:

pi@raspberrypi:~ $ if true ||; ! true; then echo a; else echo b; fi
bash: syntax error near unexpected token `;'

The question is, is this behavior a bug or is it done intentionally? And if this is done on purpose, then what is the meaning of such differences?

Answer 1

Let's say the overall syntax of if is:

if
     list-of-commands
then
     another-list-of-commands
fi

If list-of-commands exits with zero exit status, then another-list-of-commands is executed. Zero exit status is also called a "success", whereas non-zero is, well, called a "failure".

The overall rule of thumb, is that the exit status of any list of commands is the exit status of the last command executed.

The "list-of-commands" can contain multiple list-of-commands "inside" and multiple commands inside. You can put whole scripts in there.

if true or ! true; then

true is a command that exits with zero exit status. The executable true is executed with 3 arguments - string or, string ! and string true. The command true ignores arguments and exits with zero exit status.

Note that or is just or the string or it has no special meaning.

if true -o ! true; then

Same as above, just second argument is the string -o. Once again -o are characters -o.

if true || ! true; then

|| is special. || executes the command on the right side when the command on the left side exits with non-zero exit status. The exit status of || is the exit status of the command last executed.

true exits with zero exit status. || does not execute the right side. The last command is true and exited with zero exit status. The exit status of the list true || ! true is zero.

  if true or 
  ! true; then

Formatting:

if 
   true or 
   ! true
then

There are two commands here.

First true command is executed with one argument or. Its exit status is ignored.

Then true command is executed, it exits with zero exit status. ! "inverts" the exit status, so the exit status of ! true is one.

The exit status of the list of commands true or ; ! true is the exit status of the last command executed. ! true was last, it exited with non-zero. Execution is goes to the else branch.

if true -o 
 ! true; then

As above.

if true || 
 ! true; then

Empty lines or with only whitespaces and only a comment after || are ignored. This is exactly the same as the if true || ! true; then case.

if 
   true || 
   # yay a comment here

   # ^^ and empty lines too!
   ! true
 

then

if true ||; ! true; then

The syntax is command || command. ; ends a list of commands. There is no command after ||. Thus, it's a syntax error.