Command grouping (&&, ||, ...)

Question

We are currently in the /home/student/ directory. We execute the following commands:

pwd; (ls) || { cd .. && ls student/; }  && cd student || cd / && cd ;

The commands that are executed are: pwd, ls, cd student, cd /, cd

Here is what I think:

• pwd is executed, because it's the first command

• (ls) is executed in a subshell, because the commands are seperated
  with ";"

• the code on the right of || isn't executed, since the code on the
  left of || was executed

So far everything clear, I guess. But I have no idea why other commands are executed? If someone could break it down for me, I'd appreciate it.

Answer 1

Operator precedence for && and || is strictly left-to-right.

Thus:

pwd; (ls) || { cd .. && ls student/; }  && cd student || cd / && cd ;

...is equivalent to...

pwd; { { { (ls) || { cd .. && ls student/; }; } && cd student; } || cd /; } && cd ; }

...breaking that down graphically:

pwd; {                                      # 1
       {                                    # 2
         { (ls) ||                          # 3
                   { cd .. &&               # 4
                              ls student/;  # 5
                   };                       # 6
         } && cd student;                   # 7
       } || cd /;                           # 8
     } && cd ;                              # 9

1. pwd happens unconditionally
2. (Grouping only)
3. ls happens (in a subshell) unconditionally.
4. cd .. happens if (3) failed.
5. ls student/ happens if (3) failed and (4) succeeded
6. (Grouping only)
7. cd student happens if either (3) succeeded or both (4) and (5) succeeded.
8. cd / happens if either [both (3) and one of (4) or (5) failed], or [(7) failed].
9. cd happens if (7) occurred and succeeded, or (8) occurred and succeeded.

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Using explicit grouping operators is wise to avoid confusing yourself. Avoiding writing code as hard to read as this is even wiser.