Re: Bug? Explanation??

On Dec 30, 2016 11:20 PM, "Peter & Kelly Passchier" < peterke...@passchier.net> wrote: Thanks Dennis and Grisha! I understand. I had thought that every line that is piped into bash is it's own shell alignment, but I understand it's more like sourcing, so these would be more or less equivalent,

Re: Bug? Explanation??

I guess they are not equivalent, piping into bash opens a subshell, and sourcing doesn't; these act differently: echo exit |bash echo exit >r; source r Hope I've gotten it right now. Thanks, Peter On 31/12/2559 11:20, Peter & Kelly Passchier wrote: > Thanks Dennis and Grisha! I understand. >

Re: Bug? Explanation??

Thanks Dennis and Grisha! I understand. I had thought that every line that is piped into bash is it's own shell alignment, but I understand it's more like sourcing, so these would be more or less equivalent, right? r=23; echo $r; echo -e 'r=bc\necho $r' >r; source r r=23; echo $r; echo -e

Re: Bug? Explanation??

On Fri, Dec 30, 2016 at 11:03 PM, PePa wrote: r=23; echo $r; echo -e "r=bc\necho $r" |bash You need to escape the $ in the second echo statement so that $r is not evaluated to 23 before being echoed. ​

Bug? Explanation??

It seems that when piping into bash, variables have different 'retention' than functions: r=23; echo $r; echo -e "r=bc\necho $r" |bash 23 23 r(){ echo Hey;}; r; echo -e "r(){ echo Ho;}\nr" |bash Hey Ho Is this a bug, or is there a rationale? Thanks, Peter