Multiple logical operators, ((A || B) && C), and "syntax error near unexpected token"

Question

I'm working with Bash 3, and I'm trying to form a conditional. In C/C++, its dead simple: ((A || B) && C). In Bash, its turning out not so (I think the Git authors must have contributed this code before they moved onto other endeavors).

This does not work. Note that <0 or 1> is not a string literal; it means a 0 or 1 (generally comes from grep -i).

A=<0 or 1>
B=<0 or 1>
C=<0 or 1>
if [ [ "$A" -eq "0" ] || [ "$B" -ne "0" ] ] && [ "$C" -eqe "0" ]; then ... fi

It results in:

line 322: syntax error near unexpected token `[['

I then tried:

A=<0 or 1>
B=<0 or 1>
C=<0 or 1>
if [ ([ "$A" -eq "0" ]) || ([ "$B" -ne "0" ]) ] && [ "$C" -eq "0" ]; then ... fi

it results in:

line 322: syntax error near unexpected token `[['

Part of the problem is search results are the trivial examples, and not the more complex examples with compound conditionals.

How do I perform a simple ((A || B) && C) in Bash?

---

I'm ready to just unroll it and repeat the same commands in multiple blocks:

A=<0 or 1>
B=<0 or 1>
C=<0 or 1>

if [ "$A" -eq "0" ] && [ "$C" -eq "0" ]; then
    ...
elif [ "$B" -ne "0" ] && [ "$C" -eq "0" ]; then
    ... 
fi

Answer 1

The syntax of bash is not C-like, even if a little part of it is inspired by C. You can't simply try to write C code and expect it to work.

The main point of a shell is to run commands. The open-bracket command [ is a command, which performs a single test¹. You can even write it as test (without the final closing bracket). The || and && operators are shell operators, they combine commands, not tests.

So when you write

[ [ "$A" -eq "0" ] || [ "$B" -ne "0" ] ] && [ "$C" -eq "0" ]

that's parsed as

[ [ "$A" -eq "0" ] ||
[ "$B" -ne "0" ] ] &&
[ "$C" -eq "0" ]

which is the same as

test [ "$A" -eq "0" ||
test "$B" -ne "0" ] &&
test "$C" -eq "0"

Notice the unbalanced brackets? Yeah, that's not good. Your attempt with parentheses has the same problem: spurious brackets.

The syntax to group commands together is braces. The way braces are parsed requires a complete command before them, so you'll need to terminate the command inside the braces with a newline or semicolon.

if { [ "$A" -eq "0" ] || [ "$B" -ne "0" ]; } && [ "$C" -eq "0" ]; then …

There's an alternative way which is to use double brackets. Unlike single brackets, double brackets are special shell syntax. They delimit conditional expressions. Inside double brackets, you can use parentheses and operators like && and ||. Since the double brackets are shell syntax, the shell knows that when these operators are inside brackets, they're part of the conditional expression syntax, not part of the ordinary shell command syntax.

if [[ ($A -eq 0 || $B -ne 0) && $C -eq 0 ]]; then …

If all of your tests are numerical, there's yet another way, which delimit artihmetic expressions. Arithmetic expressions perform integer computations with a very C-like syntax.

if (((A == 0 || B != 0) && C == 0)); then …

You may find my bash bracket primer useful.

[ can be used in plain sh. [[ and (( are specific to bash (and ksh and zsh).

¹ _{It can also combine multiple tests with boolean operators, but this is cumbersome to use and has subtle pitfalls so I won't explain it.}

Answer 2

Use [[:

if [[ ( "$A" -eq "0" || "$B" -ne "0" ) && "$C" -eq "0" ]]; then ...

If you prefer [, the following works:

if [ \( "$A" -eq "0" -o "$B" -ne "0" \) -a "$C" -eq "0" ]; then ...

Answer 3

Use the -o operator instead of your nested ||. You can also make use of -a to replace && if needed in your other statements.

   EXPRESSION1 -a EXPRESSION2
          both EXPRESSION1 and EXPRESSION2 are true

   EXPRESSION1 -o EXPRESSION2
          either EXPRESSION1 or EXPRESSION2 is true


if [  "$A" -eq "0" -o "$B" -ne "0"  ] && [ "$C" -eq "0" ]; then echo success;fi