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Also see [[http://wiki.bash-hackers.org/syntax/arith_expr|the Bash hackers article]] about the full syntax theory. /!\ The {{{$[ ]}}} syntax is [[http://wiki.bash-hackers.org/syntax/expansion/arith|deprecated]] There are several ways to tell Bash to treat numbers as integers instead of strings, and to do basic arithmetic operations on them. The first is to use the {{{let}}} command: |
/!\ The `$[ ]` syntax is [[http://wiki.bash-hackers.org/syntax/expansion/arith|deprecated]]. Please use `$(( ))` instead. == Arithmetic Expansion == POSIX sh (and all shells based on it, including Bash and ksh) uses the `$(( ))` syntax to do arithmetic, using the same syntax as C. (See [[http://wiki.bash-hackers.org/syntax/arith_expr|the Bash hackers article]] for the full syntax.) Bash calls this an "Arithmetic Expansion", and it obeys the same basic rules as all other `$...` substitutions. `$(( ))` is the first example of a ''math context'', meaning a context where the syntax and semantics of C's integer arithmetic are used. This will be discussed in more detail below. Here are a few examples using `$(( ))`: {{{ # POSIX sh i=$((j + 3)) lvcreate -L "$((24 * 1024))" -n lv99 vg99 q=$((29 / 6)) r=$((29 % 6)) if test "$((a%4))" = 0; then ... }}} Notes: * Arithmetic operators include all of the C operators (arithmetic, bit shifting/masking, ternary `?:`), plus `**` for exponentiation. * You may use commas to separate multiple expressions within a single math context. The final ''value'' of the arithmetic expression is that of the last comma-delimited expression. * In bash, all arithmetic is done with signed integers using C's `intmax_t` variable type (typically 64 bits, but platform-dependent). Before bash 2.05b, it used `long int` variables (typically 32 bits). * Variable names in a math context are substituted with their values, as long as those values are valid integers. In Bash, this substitution continues recursively until an integer or an invalid variable name is generated. Unset or empty variables are evaluated as 0. * Numbers without leading 0 are treated as base 10. Numbers with a leading 0x are treated as base 16. Numbers with a leading 0 (not followed by x) are treated as base 8. You may put `10#` or `16#` (etc.) in front of a number to force it to be interpreted in a given base -- more on this later. == Arithmetic Commands == Bash also offers two forms of ''commands'' that use a math context. The expressions within the math context follow the same rules as the expressions inside `$(( ))`, but since we have a command, we also get an exit status, and side effects. The exit status is based on the value of the (last) expression. If the expression evaluates to 0, the command is considered "false" and returns 1. Otherwise, the command is considered "true" and returns 0. See below for examples of this. The first arithmetic command is `let`: |
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Note that as with any variable assignment in Bash, you need to quote the whole right hand side if it contains spaces, thus: | Note that each arithmetic expression has to be passed as a single argument to the `let` command, so you need quotes if there are spaces or globbing characters. Thus: |
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}}} Division in Bash is integer division, and it truncates the results, just as in C: {{{ let a=28/6 echo "a = $a" # Prints a = 4 }}} In addition to the {{{let}}} command, one may use the {{{(( ))}}} syntax to enforce an arithmetic context. If there is a {{{$}}} (dollar sign) before the parentheses, then a substitution is performed (more on this below). White space is allowed inside {{{(( ))}}} with much greater leniency than with normal assignments, and variables inside {{{(( ))}}} don't require {{{$}}} (because string literals aren't allowed). Examples: |
let 'a = 17 + 23' # Right let a=17 a+=23 # Right (2 arithmetic expressions) let a[1]=1+1 # Wrong (try after touch a1=1+1 or with shopt -s failglob) let 'a[1]=1+1' # Right let a\[1\]=1+1 # Right, but very odd. }}} The second command is `(( ))`. It works identically to `let`, but you don't need to quote the expressions because they are delimited by the `((` and `))` syntax. For example: |
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# (( )) may also be used as a command. > or < inside (( )) means # greater/less than, not output/input redirection. |
}}} `>` or `<` inside `(( ))` means greater/less than, not output/input redirection. An expression containing one of these evalutes to either 1 (if the comparison is true) or 0 (if false). {{{ |
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{{{(( ))}}} without the leading {{{$}}} is not a standard sh feature. It comes from ksh and is only available in ksh, Bash and zsh. {{{$(( ))}}} substitution is allowed in the POSIX shell. As one would expect, the result of the arithmetic expression inside the {{{$(( ))}}} is substituted into the original command. Like for parameter substitution, arithmetic substitution is subject to work splitting so should be quoted to prevent it when in list contexts. Here are some examples of the use of the arithmetic substitution syntax: {{{ a=$((a+7)) # POSIX-compatible version of previous code. if test "$((a%4))" = 0; then ... lvcreate -L "$((4*1096))" -n lvname vgname # Actual HP-UX example. }}} Variables may be {{{declare}}}d as integers so that any subsequent assignments to them will always assume a numeric context. Essentially any variable that's declared as an integer acts as if you had a {{{let}}} command in front of it when you assign to it. For example: {{{ unset b # Forget any previous declarations b=7+5; echo "$b" # Prints 7+5 declare -i b # Declare b as an integer b=7+5; echo "$b" # Prints 12 }}} Also, array indices are a numeric context: {{{ n=0 while read line; do array[n++]=$line # array[] forces a numeric context done }}} There is one common pitfall with arithmetic expressions in Bash: numbers with leading zeroes are treated as octal. For example, {{{ # Suppose today is September 19th. month=$(date +%m) next_month=$(( (month == 12) ? 1 : month+1 )) # bash: 09: value too great for base (error token is "09") }}} This causes great confusion among people who are extracting zero-padded numbers from various sources (although dates are by far the most common) and then doing math on them without sanitizing them first. (It's especially bad if you write a program like this in March, test it, roll it out... and then it doesn't blow up until August 1.) If you have leading-zero problems with Bash's built-in arithmetic, there are two possible solutions. The first is, obviously, to remove the leading zeroes from the numbers before doing math with them. This is not trivial in Bash, unfortunately, because Bash has no ability to perform substitutions on a variable using regular expressions (it can only do it with "glob" patterns). But you could use a loop: {{{ # This removes leading zeroes from a, one at a time. while [[ $a = 0* ]]; do a=${a#0}; done }}} You can do the above without using a loop, by using extended globs; see [[BashFAQ/067|FAQ #67]] for more information. Or, you could use {{{sed}}}; that may be more efficient if you're reading many numbers from a stream, and can arrange to sanitize them all in one command, rather than one by one. Without a loop: {{{ # This removes leading zeroes from a, all at once. a=${a##+(0)} }}} The third solution is to force Bash to treat all numbers as base 10 by prefixing them with {{{10#}}}. This might be more efficient, but also may be less elegant to read. {{{ a=008 let b=a+1 # Generates an error because 008 is not valid in octal. let b=10#$a+1 # Force a to be treated as base 10. Note: the $ is required. }}} Finally, a note on the exit status of commands, and the notions of "true" and "false", is in order. When bash runs a command, that command will return an exit status from 0 to 255. 0 is considered "success" (which is "true" when used in the context of an {{{if}}} or {{{while}}} command). However, in an ''arithmetic'' context, there are places where the C language rules (0 is false, anything else is true) apply. |
`(( ))` is used more widely than `let`, because it fits so well into an `if` or `while` command. Finally, a note on the exit status of commands, and the notions of "true" and "false", is in order. When Bash runs a command, that command will return an exit status from 0 to 255. 0 is considered "success" (which is "true" when used in the context of an `if` or `while` command). However, when evaluating an arithmetic expression, C language rules (0 is false, anything else is true) apply. |
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echo "$((10 > 6))" # Writes 1. An arithmetic expression returns 1 for true. }}} In addition to a comparison returning 1 for true, an arithmetic expression that ''evaluates'' to a non-zero value is also true in the sense of a command. |
echo "$((10 > 6))" # Writes 1. An arithmetic expression evaluates to 1 for true. }}} In addition to a comparison returning 1 for true, an arithmetic command that ''evaluates'' to any non-zero value returns true as an exit status. |
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This also means that every arithmetic command we run is returning an exit status. Most scripts ignore these, but if you are using [[BashFAQ/105|set -e]] you may be unpleasantly surprised when your program ''aborts'' because you assigned 0 to a number. == Math Contexts == As we've seen, the insides of `$(( ))` and `(( ))`, and the arguments of `let`, are math contexts. Also, (non-associative) [[BashFAQ/005|array]] indices are a math context: {{{ n=0 while read line; do array[n++]=$line # array[] forces a numeric context done }}} In case it wasn't obvious, the `(( ))` in a C-style `for` command are a math context. Or three separate math contexts, depending on your point of view. {{{ for ((i=0, j=0; i<100; i++, j+=5)); do ... }}} Finally, the ''start'' and ''length'' arguments of Bash's `${var:start:length}` [[BashFAQ/073|parameter expansion]] are math contexts. {{{ echo "${string:i+2:len-2}" }}} == Leading Zeros and Base Selection == There is one common pitfall with arithmetic expressions in Bash: numbers with leading zeroes are treated as octal. For example, {{{ # Suppose today is September 19th. month=$(date +%m) next_month=$(( (month == 12) ? 1 : month+1 )) # bash: 09: value too great for base (error token is "09") }}} This causes great confusion among people who are extracting zero-padded numbers from various sources (dates and times are by far the most common) and then doing math on them without sanitizing them first. It's especially bad if you write a program like this in March, test it, roll it out... and then it doesn't blow up until August 1. There are two possible solutions. The first is, obviously, to remove the leading zeroes from the numbers before doing math with them. This can be done in several ways. First, a simple loop: {{{ # This removes leading zeroes from a, one at a time. while [[ $a = 0* ]]; do a=${a#0}; done }}} You can also do it with [[glob|extended globs]]; see [[BashFAQ/067|FAQ #67]] for more information. Or, you could use {{{sed}}}; that may be more efficient if you're reading many numbers from a stream, and can arrange to sanitize them all in one command, rather than one by one. With an extended glob: {{{ shopt -s extglob # This removes leading zeroes from a, all at once. a=${a##+(0)} }}} Finally, Bash can do it with an arithmetic expression: {{{ month=$(date +%m) month=$((10#$month)) # Strip leading zeros by forcing evaluation in base 10 }}} The second solution is to force Bash to treat all numbers as base 10 by prefixing them with `10#` every time they are used. {{{ a=008 let b=a+1 # Generates an error because 008 is not valid in octal. let b=10#$a+1 # Force a to be treated as base 10. Note: the $ is required. }}} Generally, it's better to strip the zeroes once than to keep doing `10#` on the same value repeatedly. |
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todec() { | frombase() { |
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}}} Examples: {{{ todec 16 ffe # -> 4094 todec 2 100100 # -> 36 }}} |
# Examples: frombase 16 ffe # prints 4094 frombase 2 100100 # prints 36 }}} == Integer Declaration == /!\ '''Integer declaration is considered harmful.''' Don't do this. It makes your code difficult to read, because you need to know whether a given variable is declared as an integer or not, to know what a command does. Variables may be {{{declare}}}d as integers so that any subsequent assignments to them will always assume a numeric context. Essentially any variable that's declared as an integer acts as if you had a {{{let}}} command in front of it when you assign to it. For example: {{{ unset b # Forget any previous declarations b=7+5; echo "$b" # Prints 7+5 declare -i b # Declare b as an integer b=7+5; echo "$b" # Prints 12 }}} Also note that regardless of whether you use `declare -i`, bash '''does not''' store numbers in binary form. They are still stored as strings, with string/integer conversions happening on the fly every time arithmetic is done. |
Arithmetic in BASH is integer math only. You can't do floating point math in Bash; if you need that capability, see Bash FAQ #22.
The $[ ] syntax is deprecated. Please use $(( )) instead.
Arithmetic Expansion
POSIX sh (and all shells based on it, including Bash and ksh) uses the $(( )) syntax to do arithmetic, using the same syntax as C. (See the Bash hackers article for the full syntax.) Bash calls this an "Arithmetic Expansion", and it obeys the same basic rules as all other $... substitutions.
$(( )) is the first example of a math context, meaning a context where the syntax and semantics of C's integer arithmetic are used. This will be discussed in more detail below.
Here are a few examples using $(( )):
# POSIX sh i=$((j + 3)) lvcreate -L "$((24 * 1024))" -n lv99 vg99 q=$((29 / 6)) r=$((29 % 6)) if test "$((a%4))" = 0; then ...
Notes:
Arithmetic operators include all of the C operators (arithmetic, bit shifting/masking, ternary ?:), plus ** for exponentiation.
You may use commas to separate multiple expressions within a single math context. The final value of the arithmetic expression is that of the last comma-delimited expression.
In bash, all arithmetic is done with signed integers using C's intmax_t variable type (typically 64 bits, but platform-dependent). Before bash 2.05b, it used long int variables (typically 32 bits).
- Variable names in a math context are substituted with their values, as long as those values are valid integers. In Bash, this substitution continues recursively until an integer or an invalid variable name is generated. Unset or empty variables are evaluated as 0.
Numbers without leading 0 are treated as base 10. Numbers with a leading 0x are treated as base 16. Numbers with a leading 0 (not followed by x) are treated as base 8. You may put 10# or 16# (etc.) in front of a number to force it to be interpreted in a given base -- more on this later.
Arithmetic Commands
Bash also offers two forms of commands that use a math context. The expressions within the math context follow the same rules as the expressions inside $(( )), but since we have a command, we also get an exit status, and side effects.
The exit status is based on the value of the (last) expression. If the expression evaluates to 0, the command is considered "false" and returns 1. Otherwise, the command is considered "true" and returns 0. See below for examples of this.
The first arithmetic command is let:
let a=17+23 echo "a = $a" # Prints a = 40
Note that each arithmetic expression has to be passed as a single argument to the let command, so you need quotes if there are spaces or globbing characters. Thus:
let a=17 + 23 # WRONG let a="17 + 23" # Right let 'a = 17 + 23' # Right let a=17 a+=23 # Right (2 arithmetic expressions) let a[1]=1+1 # Wrong (try after touch a1=1+1 or with shopt -s failglob) let 'a[1]=1+1' # Right let a\[1\]=1+1 # Right, but very odd.
The second command is (( )). It works identically to let, but you don't need to quote the expressions because they are delimited by the (( and )) syntax. For example:
((a=$a+7)) # Add 7 to a ((a = a + 7)) # Add 7 to a. Identical to the previous command. ((a += 7)) # Add 7 to a. Identical to the previous command. ((a = RANDOM % 10 + 1)) # Choose a random number from 1 to 10. # % is modulus, as in C.
> or < inside (( )) means greater/less than, not output/input redirection. An expression containing one of these evalutes to either 1 (if the comparison is true) or 0 (if false).
if ((a > 5)); then echo "a is more than 5"; fi
(( )) is used more widely than let, because it fits so well into an if or while command.
Finally, a note on the exit status of commands, and the notions of "true" and "false", is in order. When Bash runs a command, that command will return an exit status from 0 to 255. 0 is considered "success" (which is "true" when used in the context of an if or while command). However, when evaluating an arithmetic expression, C language rules (0 is false, anything else is true) apply.
Some examples:
true; echo "$?" # Writes 0, because a successful command returns 0. ((10 > 6)); echo "$?" # Also 0. An arithmetic command returns 0 for true. echo "$((10 > 6))" # Writes 1. An arithmetic expression evaluates to 1 for true.
In addition to a comparison returning 1 for true, an arithmetic command that evaluates to any non-zero value returns true as an exit status.
if ((1)); then echo true; fi # Writes true.
This also lets you use "flag" variables, just like in a C program:
found=0 while ...; do ... if something; then found=1; fi # Found one! Keep going. ... done if ((found)); then ...
This also means that every arithmetic command we run is returning an exit status. Most scripts ignore these, but if you are using set -e you may be unpleasantly surprised when your program aborts because you assigned 0 to a number.
Math Contexts
As we've seen, the insides of $(( )) and (( )), and the arguments of let, are math contexts.
Also, (non-associative) array indices are a math context:
n=0 while read line; do array[n++]=$line # array[] forces a numeric context done
In case it wasn't obvious, the (( )) in a C-style for command are a math context. Or three separate math contexts, depending on your point of view.
for ((i=0, j=0; i<100; i++, j+=5)); do ...
Finally, the start and length arguments of Bash's ${var:start:length} parameter expansion are math contexts.
echo "${string:i+2:len-2}"
Leading Zeros and Base Selection
There is one common pitfall with arithmetic expressions in Bash: numbers with leading zeroes are treated as octal. For example,
# Suppose today is September 19th. month=$(date +%m) next_month=$(( (month == 12) ? 1 : month+1 )) # bash: 09: value too great for base (error token is "09")
This causes great confusion among people who are extracting zero-padded numbers from various sources (dates and times are by far the most common) and then doing math on them without sanitizing them first. It's especially bad if you write a program like this in March, test it, roll it out... and then it doesn't blow up until August 1.
There are two possible solutions. The first is, obviously, to remove the leading zeroes from the numbers before doing math with them. This can be done in several ways. First, a simple loop:
# This removes leading zeroes from a, one at a time. while [[ $a = 0* ]]; do a=${a#0}; done
You can also do it with extended globs; see FAQ #67 for more information. Or, you could use sed; that may be more efficient if you're reading many numbers from a stream, and can arrange to sanitize them all in one command, rather than one by one.
With an extended glob:
shopt -s extglob # This removes leading zeroes from a, all at once. a=${a##+(0)}
Finally, Bash can do it with an arithmetic expression:
month=$(date +%m) month=$((10#$month)) # Strip leading zeros by forcing evaluation in base 10
The second solution is to force Bash to treat all numbers as base 10 by prefixing them with 10# every time they are used.
a=008 let b=a+1 # Generates an error because 008 is not valid in octal. let b=10#$a+1 # Force a to be treated as base 10. Note: the $ is required.
Generally, it's better to strip the zeroes once than to keep doing 10# on the same value repeatedly.
Here is a function to convert numbers in other bases to decimal (base 10):
frombase() { echo "$(( $1#$2 ))" } # Examples: frombase 16 ffe # prints 4094 frombase 2 100100 # prints 36
Integer Declaration
Integer declaration is considered harmful. Don't do this. It makes your code difficult to read, because you need to know whether a given variable is declared as an integer or not, to know what a command does.
Variables may be declared as integers so that any subsequent assignments to them will always assume a numeric context. Essentially any variable that's declared as an integer acts as if you had a let command in front of it when you assign to it. For example:
unset b # Forget any previous declarations b=7+5; echo "$b" # Prints 7+5 declare -i b # Declare b as an integer b=7+5; echo "$b" # Prints 12
Also note that regardless of whether you use declare -i, bash does not store numbers in binary form. They are still stored as strings, with string/integer conversions happening on the fly every time arithmetic is done.