Parenthesis in expr arithmetic: 3 * (2 + 1)
Solution 1
Another way to use let
bash builtin:
$ let a="3 * (2 + 1)"
$ printf '%s\n' "$a"
9
Note
As @Stéphane Chazelas pointed out, in bash
you should use ((...))
to do arithmetic over expr
or let
for legibility.
For portability, use $((...))
like @Bernhard answer.
Solution 2
You can use the arithmetic expansion instead.
echo "$(( 3 * ( 2 + 1 ) ))"
9
In my personal opinion, this looks a bit nicer than using expr
.
From man bash
Arithmetic Expansion Arithmetic expansion allows the evaluation of an arithmetic expression and the substitution of the result. The format for arithmetic expansion is:
$((expression))
The expression is treated as if it were within double quotes, but a double quote inside the parentheses is not treated specially. All tokens in the expression undergo parameter expansion, string expansion, command substitution, and quote removal. Arithmetic expansions may be nested.
The evaluation is performed according to the rules listed below under ARITHMETIC EVALUATION. If expression is invalid, bash prints a message indicating failure and no substitution occurs.
Solution 3
There's no reason to be using expr
for arithmetic in modern shells.
POSIX defines the $((...))
expansion operator. So you can use that in all POSIX compliant shells (the sh
of all modern Unix-likes, dash, bash, yash, mksh, zsh, posh, ksh...).
a=$(( 3 * (2 + 1) ))
a=$((3*(2+1)))
ksh
also introduced a let
builtin which is passed the same kind of arithmetic expression, doesn't expand into something but returns an exit status based on whether the expression resolves to 0 or not, like in expr
:
if let 'a = 3 * (2 + 1)'; then
echo "$a is non-zero"
fi
However, as the quoting makes it awkward and not very legible (not to the same extent as expr
of course), ksh
also introduced a ((...))
alternative form:
if (( a = 3 * (2 + 1) )) && (( 3 > 1 )); then
echo "$a is non-zero and 3 > 1"
fi
((a+=2))
which is a lot more legible and should be used instead.
let
and ((...))
are only available in ksh
, zsh
and bash
. The $((...))
syntax should be preferred if portability to other shells is needed, expr
is only needed for pre-POSIX Bourne-like shells (typically the Bourne shell or early versions of the Almquist shell).
On the non-Bourne front, there are a few shells with built-in arithmetic operator:
-
csh
/tcsh
(actually the first Unix shell with arithmetic evaluation built-in):@ a = 3 * (2 + 1)
-
akanga
(based onrc
)a = $:'3 * (2 + 1)'
as a history note, the original version of the Almquist shell, as posted on usenet in 1989 had an
expr
builtin (actually merged withtest
), but it was removed later.
Solution 4
expr
is an external command, it is not special shell syntax. Therefore, if you want expr
to see shell special characters, you need to protect them from shell parsing by quoting them. Furthermore, expr
needs each number and operator to be passed as a separate parameter. Thus:
expr 3 \* \( 2 + 1 \)
Unless you're working on an antique unix system from the 1970s or 1980s, there is very little reason to use expr
. In the old days, shells didn't have a built-in way to perform arithmetic, and you had to call the expr
utility instead. All POSIX shells have built-in arithmetic via the arithmetic expansion syntax.
echo "$((3 * (2 + 1)))"
The construct $((…))
expands to the result of the arithmetic expression (written in decimal). Bash, like most shells, supports only integer arithmetic modulo 264 (or modulo 232 for older versions of bash and some other shells on 32-bit machines).
Bash offers an additional convenience syntax when you want to perform assignments or to test whether an expression is 0 but don't care about the result. This construct also exists in ksh and zsh but not in plain sh.
((x = 3 * (2+1)))
echo "$x"
if ((x > 3)); then …
In addition to integer arithmetic, expr
offers a few string manipulation functions. These too are subsumed by features of POSIX shells, except for one: expr STRING : REGEXP
tests whether the string matches the specified regexp. A POSIX shell cannot do this without external tools, but bash can with [[ STRING =~ REGEXP ]]
(with a different regexp syntax — expr
is a classic tool and uses BRE, bash uses ERE).
Unless you're maintaing scripts that run on 20-year-old systems, you don't need to know that expr
ever existed. Use shell arithmetic.
Solution 5
Use parenthesis with quotes:
expr 3 '*' '(' 2 '+' 1 ')'
9
The quotes prevent bash from interpreting the parenthesis as bash syntax.
Related videos on Youtube
Vitalij
I am Nicolas Raoul, IT consultant in Tokyo. Feel free to copy/paste the source code from my StackExchange answers, I release it to the public domain.
Updated on September 18, 2022Comments
-
Vitalij almost 2 years
expr
does not seem to like parenthesis (used in mathematics to explicit operator priority):expr 3 * (2 + 1) bash: syntax error near unexpected token `('
How to express operator priority in bash?
-
Vitalij almost 10 years+1 Even more readable! I posted my question+answer just thinking it would be helpful for my fellow Linux users, but now I am getting a lot of benefit from the other answers :-)
-
chepner almost 10 yearsAside from readability, it also doesn't require forking an extra process to do the arithmetic; it's handled by the shell itself.
-
Stéphane Chazelas almost 10 yearsThere's no reason to be using
let
. It's not any more standard or portable than(( a = 3 * (2 + 1) ))
(both come fromksh
and are only available in ksh, bash and zsh) and it's less legible or easy to quote. Usea=$((3 * (2 + 1)))
to be portable. -
Stéphane Chazelas almost 10 yearsI'm not saying it's wrong, I'm just saying it should not be used as there are better alternatives (one for legibility
((a = 3 * (2 + 1) ))
, one for portabilitya=$((3 * (2 + 1)))
), so it's not a note against you or your answer but against it being the selected answer and top-scorer. -
Stéphane Chazelas almost 10 yearsNote that in POSIX shells, it's subject to word splitting, so it's a good habit to quote it in list contexts.
-
cuonglm almost 10 years@StéphaneChazelas: Updated my answer!
-
G-Man Says 'Reinstate Monica' almost 10 yearsWhat Nicolas illustrates but doesn’t explain is that the tokens on the
expr
command line must be separated by spaces; so; for example,expr 3 "*" "(2" "+" "1)"
will not work. (Also, BTW, you probably don’t need to quote the+
.) -
Stéphane Chazelas almost 10 years
expr foo : '\(.\)'
also does text extraction.bash
'sBASH_REMATCH
achieves something similar. It also does string comparison, which POSIX[
does not do (though one could imagine ways to usesort
for that). -
MattBianco almost 10 yearsI learn something new every day from you, Stéphane. I very much appreciate your POSIX shell knowledge!
-
Gordon about 9 yearsI have always used
a=1 $[a+2]
ora=1 b=2 $[a+b]
. Is their reason to avoid that syntax? -
Mingye Wang over 8 yearsHow about
: $((a = a*2))
? -
Sushant over 8 yearsunderscore as syntax placeholder --- you a Schemer, @Giles ? :]
-
Gilles 'SO- stop being evil' over 8 years@RubyTuesdayDONO I didn't use an underscore here. Are you misreading U+2026 HORIZONTAL ELLIPSIS? If so, try using a bigger font.
-
Sushant over 8 years@Giles — okay, yes, it only looks like an underscore because of my font size. for me, "Schemer" is a complement, and it's not like ellipsis versus underscore changes the meaning anyway … no need to get snarky over it :/
-
lordhog almost 8 yearsWhen I try this at the bash shell I get 'Illegal variable name."
-
Simon Lindholm over 5 years@Gordon: The bash man page says "The old format $[expression] is deprecated and will be removed in upcoming versions of bash."
-
Blaise about 5 yearsWhat if I have a floating point? My expression is a=$(( -14 + 0.2 * (1+2+3) )). The error token is ".2*(1+2+3)"
-
Stéphane Chazelas about 5 years@Blaise, then you'd need a shell that supports floating points in
$((...))
like zsh, ksh93 or yash. -
hackerb9 almost 3 years@Blaise If you need floating point, you can just use an external program, like so
a=$(bc <<<"-14 + 0.2 * (1+2+3)")