| 1 |
On Monday 17 June 2013 01:42:15 Mike Frysinger wrote: |
| 2 |
> On Sunday 02 June 2013 13:38:04 Steven J. Long wrote: |
| 3 |
> > On Sat, Jun 01, 2013 at 11:03:20PM -0400, Mike Frysinger wrote: |
| 4 |
> > > +# is not specified, the var will be unset. |
| 5 |
> > > +evar_push_set() { |
| 6 |
> > > + local var=$1 |
| 7 |
> > > + evar_push ${var} |
| 8 |
> > > + case $# in |
| 9 |
> > > + 1) unset ${var} ;; |
| 10 |
> > > + 2) eval ${var}=\$2 ;; |
| 11 |
> > |
| 12 |
> > I wish you wouldn't use eval for this. I know it's technically okay here, |
| 13 |
> > or would be if you verified the parameter, but bash has printf -v for |
| 14 |
> > this purpose: |
| 15 |
> |
| 16 |
> interesting, i hadn't seen that before ... looks new to bash-3.1. /me |
| 17 |
> tucks that into his tool belt. |
| 18 |
> |
| 19 |
> although it doesn't quite work in the edge case where the value is an empty |
| 20 |
> string. consider: |
| 21 |
> unset x |
| 22 |
> printf -v x '' |
| 23 |
> echo ${x+set} |
| 24 |
> |
| 25 |
> that should show "set", but it does not. i'll have to keep `eval ${var}=` |
| 26 |
> when the value we're setting is empty. or just keep the eval code since i |
| 27 |
> have to do eval anyways at that point. |
| 28 |
> |
| 29 |
> i'll report it upstream to the bash guys. |
| 30 |
|
| 31 |
looks like it can be worked around by doing: |
| 32 |
printf -v x '%s' '' |
| 33 |
which is arguably what we want anyways |
| 34 |
-mike |
Archives