incrementing within a shell script?

[spiderpig]

I know that the following:

#!/bin/sh
i=0
i=$(($i + 1))
echo $i

is the same as the following:

#!/bin/sh
i=0
i=$(expr $i + 1)
echo $i

My understanding is that a new shell is spawned to execute ($i + 1) or expr $i + 1, but why are the parentheses needed in ($i + 1)? The best answer I can figure is that the statement:

i=$($i + 1)

is syntactically wrong because somehow $i in the spawned shell is being interpreted as the name of some application which is to be run. Adding the inner set of parentheses somehow forces the shell to treat the argument ($i + 1) as an expression to be evaluated. I had convinced myself that this was a plausible answer for quite some time until I tried it manually at a shell prompt:

$ i=1
$ echo $i
1
$ sh ($i + 1)
sh: 1: No such file or directory

So why are the inner set of parentheses required in x=$(($i + 1))?

Thanks!

[TerryP]

() executes a subshell
$FOO = get replace $FOO with value of FOO, so
$() executes a subshell and replaces $() with value of subshell

The `$' character is used to introduce parameter expansion, command substitution, or arithmetic evaluation.

Thus since (cmds) is like doing sh -c "cmds", $(cmds) has to substitute the value of cmds, hence $(cmds) evaluates to the output of 'cmds'. If $ triggers expansion/substitution, that makes a lot of sense right? To my understanding, in the old days: the only way to do simple arithmetic like was with expr. Then later $((expression)) was added to the language for convenience, and people could stop cursing if they missed a space (e.g. expr 2 +2) and had to retype the entire line again.



I can tell you this, shells/v7sh does not allow the $((expression)) syntax that modern /bin/sh's do. Maybe someone like DrJ could explain the history $(()) and expr, I've only used BSD for ~3 years... So I can't really say a lot that is concrete!

[spiderpig]

Originally Posted by TerryP

Then later $((expression)) was added to the language for convenience, ...

You have not addressed how the nested set of parentheses are parsed. This is heart of my question which has not been answered.

[ephemera]

$(command) does command substitution

$((expression)) does arithmetic expansion

please see the sh(1) man page.

i=5 ; j=$($i + 1) is not syntactically wrong. here the parameter expansion is done for the token $i then the command 5 is run with the arguments + and 1 and its standard o/p is assigned to $j. (obviously, there is no command by the name 5 on the system.)

i=5 ; j=$(($i + 1)) is what you want. arithmetic evaluation of the expression inside $((...)) is performed and the result 6 is assigned to j.

> how the nested set of parentheses are parsed?

observe the o/p of the following commands:

echo $((ls -l))

echo $((ls -l /)) # if you can tell why the error occurs then you have understood it :-)

echo $( (ls -l /)) # note the space used to seperate the tokens $( and (

[spiderpig]

Thanks, ephemera for answering my question! Where can this information be found? I haven't seen it anywhere.

[ephemera]

> Where can this information be found?

its all there in the man page. do check it out.

also i think we are unnecessarly complicating it. its quite simple really: $(...) and $((...)) do two different jobs and there is no connection between them other than a lexical similiarity.