Created SC1116 (markdown)

koalaman
2017-04-17 21:15:21 -07:00
parent b6c9a4f2f5
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## Missing $ on a $((..)) expression? (or use ( ( for arrays).
### Problematic code:
```sh
var=((foo+1))
```
### Correct code:
```sh
var=$((foo+1))
```
### Rationale:
You appear to be missing the `$` on an assignment from an arithmetic expression `var=$((..))` .
Without the `$`, this is an array expression which is either nested (ksh) or invalid (bash).
### Exceptions:
If you are trying to define a multidimensional Ksh array, add spaces between the `( (` to clarify:
var=( (1 2 3) (4 5 6) )