[SOLVED] Bash: syntax for ((<arithmetic expression>)) usage?

catkin · 03-18-2010, 02:24 AM

Hello

I do not understand bash' syntax regarding the use of ((<arithmetic expression>)).

$((<arithmetic expression>)) is tokenised to a single word. OK.
.
((<arithmetic expression>)) may be used in some places:
1. In array indexing
  Code:
```
array[((<arithmetic expression>))]=value
```
2. In isolation
  Code:
```
((<arithmetic expression>))
```
It may not be used where a word is expected. This generates a syntax error. Why?
Code:
```
echo ((<arithmetic expression>))
```

Best

Charles

penguiniator · 03-18-2010, 03:05 AM

Code:

echo $(( <arithmetic expression> ))

You must use a dollar sign in front of the expression where a string would be expected. Otherwise it does not send anything to standard output. For instance, you can place an arithmetic expression in place of any other command on a line by itself:

Code:

(( var++ ))

This is a valid expression. It increments var, but sends nothing to standard output. The dollar sign forces the result of the expression to standard output. So, this would output the result of incrementing var:

Code:

echo $(( ++var ))

ghostdog74 · 03-18-2010, 03:21 AM

Quote:

Originally Posted by catkin

[/LIST][*]It may not be used where a word is expected. This generates a syntax error. Why?

Code:

echo ((<arithmetic expression>))

[/LIST]Best

Charles

((i++)) increment the value of i. if you want to echo out the value, you need to put "$" in front, ie "echo $((i++))", just like any other shell variable interpolation.

colucix · 03-18-2010, 03:45 AM

Hi, Charles. I second previous posts. Just take in mind that the $ sign is the way by which the shell retrieves the value of a variable, an expression, a substitution and so on.

chrism01 · 03-18-2010, 06:48 AM

I agree with colucix; it's more accurately value retrieval, rather than sending to stdout, although the effect is pretty much the same, at least for echo.

tuxdev · 03-18-2010, 10:11 AM

Indeed, $(()) is a substitution just like ${}. (()) isn't, so it can't be used where you really need a substitution.

BTW, contexts that are already arithmetic expressions don't need the (()). They don't hurt, but what you're really doing is parenthesizing the whole expression twice. Note:

Code:

tim@epoch:~ >foo[4+5]=foo
tim@epoch:~ >echo "${foo[9]}"
foo

catkin · 03-18-2010, 10:58 AM

Thanks to all

That has really helped. It's very useful to have a clear picture of how these things work and I was stumbling around in the woods, looking for trees.

If I may test my new understanding by summarising:

((<arithmetic expression>)) is a command. It evaluates <arithmetic expression> which can be used to assign values and to increment and decrement values.
$((<arithmetic expression>)) runs the same command and the shell, while tokenising, substitutes the value of the <arithmetic expression>.
In places where an arithmetic expression is required, ((<arithmetic expression>)) may be used but there is no benefit over using <arithmetic expression>.

OK?