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I was searching for a way to convert hexadecimal via command line and found there is a very easy method echo $((0x63)).

It's working great but I'm a little confused as to what is happening here.

I know $(...) is normally a sub-shell, where the contents are evaluated before the outer command.

Is it still a sub-shell in this situation? I'm thinking not as that would mean the sub-shell is just evaluating (0x63) which isn't a command.

Can someone break down the command for me?

2 Answers 2

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$(...) is a command substitution (not just a subshell), but $((...)) is an arithmetic expansion.

When you use $((...)), the ... will be interpreted as an arithmetic expression. This means, amongst other things, that a hexadecimal string will be interpreted as a number and converted to decimal. The whole expression will then be replaced by the numeric value that the expression evaluates to.

Like parameter expansion and command substitution, $((...)) should be quoted as to not be affected by the shell's word splitting and filename globbing.

echo "$(( 0x63 ))"

As a side note, variables occurring in an arithmetic expression do not need their $:

$ x=030; y=30; z=0x30
$ echo "$(( x + y +x ))"
78
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    Also note that because $((...)) conflicts with command substitution, one has to remember to write echo "$( (echo x; echo y) | tr xy yx)" instead of echo "$((echo x; echo y) | tr xy yx)" (even though some shells will still accept the latter, YMMV). Commented Nov 1, 2017 at 14:52
  • @StéphaneChazelas And how do you make $(( )) produce output which is affected by word splitting or globbing? Commented Nov 1, 2017 at 15:25
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    @HaukeLaging, IFS=9; echo $((0x63)). See also Security implications of forgetting to quote a variable in bash/POSIX shells (What about [ $# -gt 1 ] section) Commented Nov 1, 2017 at 15:28
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This is not a subshell, but arithmetic evaluation. From man bash:

((expression))

The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to let "expression".

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    ((...)) is not quite the same as $((...)). Both does arithmetic evaluation, but the latter is replaced by the resulting value as a string.
    – Kusalananda
    Commented Nov 1, 2017 at 14:43

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