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From bash manual

Shell variables are allowed as operands; parameter expansion is performed before the expression is evaluated. Within an expression, shell variables may also be referenced by name without using the parameter expansion syntax. A shell variable that is null or unset evaluates to 0 when referenced by name without using the parameter expansion syntax. The value of a variable is evaluated as an arithmetic expression when it is referenced, or when a variable which has been given the integer attribute using ‘declare -i’ is assigned a value. A null value evaluates to 0. A shell variable need not have its integer attribute turned on to be used in an expression.

What does "The value of a variable is evaluated as an arithmetic expression when it is referenced, or when a variable which has been given the integer attribute using ‘declare -i’ is assigned a value" mean?

Is the variable assumed to be used as an operand in an arithmetic expression? var=3+4 alone will not be evaluated as an arithmetic expression.

Can you give an example for the case "when it is referenced", and an example for the case "when a variable which has been given the integer attribute using ‘declare -i’ is assigned a value"?

Thanks.

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“When it is referenced” means “when it is used”, i.e. the context indicates that the variable’s value should be interpreted as a number:

$ i=Hello
$ echo ${i}
Hello
$ echo $((i + 1))
1

Here i obviously contains a string, but because it’s used in an arithmetic expression, it’s interpreted as an arithmetic expression. If Hello is unset or null here, then it becomes 0; you can try playing around with different values for i and Hello to see the effects (i=1+1, Hello=Hello etc.).

Variables with the integer attribute, i.e. variables declared using declare -i, are coerced to integers on assignment, not on use:

$ declare -i i
$ i=Hello
$ echo ${i}
0

Note that in all these cases, no errors are reported.

Integer-attribute variables can be used with arithmetic expressions without any syntactic sugar:

$ declare -i i
$ i=3+4
$ echo ${i}
7

which is a direct consequence of the fact that their assignments are interpreted as arithmetic expressions.

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  • It's treated as 0 only if $Hello is unset or set to something that evaluates to 0. Hello is a valid arithmetic expression. If $Hello contained Hello + 1, bash would complain about the infinite recursion.
    – Stéphane Chazelas
    Jul 26 '17 at 15:28
  • Oh wow, I did not know that... So there’s an additional level of indirection!
    – Stephen Kitt
    Jul 26 '17 at 15:29
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    It's recursive. Also note that command substitutions may be run via that mechanism, so arithmetic evaluation upon uncontrolled data is a command injection vulnerability.
    – Stéphane Chazelas
    Jul 26 '17 at 15:30
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    It's more that when an arithmetic expression contains a variable name (including array members like a[2], and even (hence the ACE) a[`reboot`0]), then its content is evaluated as an arithmetic expression and the result substituted. But as a whole, it's different from using $var. For instance a=1+1; echo "$((a * 3)) $(($a * 3))" outputs 6 4. One can play with things like a=a=b++ b=1 bash -c 'echo "$((++a)) $b"'
    – Stéphane Chazelas
    Jul 26 '17 at 15:42
  • Note that in i=1+1; typeset -i i, the arithmetic expression is evaluated upon typeset in ksh93, zsh and mksh, but not in bash.
    – Stéphane Chazelas
    Jul 26 '17 at 16:24

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