Re: how to print all 3 char length string made whitespace {0..9} {a..z}

[Marco Ippolito]

On 02/08/2020 18:10, Mike Jonkmans wrote:
> On Sun, Aug 02, 2020 at 03:43:05PM -0500, Peng Yu wrote:
> > As I said, I just want to generate all three character strings with
> > space, a..z, and 0..9 in bash. Otherwise, I will have to use a
> > different programming lang which I don't want to. I don't understand
> > why you ask this question.
> >
> > On 8/2/20, Eli Schwartz <eschwartz@archlinux.org> wrote:
> > > On 8/2/20 3:44 PM, Peng Yu wrote:
> > > > Is there a good way to generate all three-character strings made of
> > > > whitespace, {0..9} and {a..z}? Thanks.
> > >
> > > You can always use multiple brace expansions together:
> > >
> > > $ printf "<%s>\n" {{a..z}," ",{0..9}}
> > >
> > > Though I'd first want to ask what you're really trying to do...
> > > (I know, I know, you'll never tell us...)
>
>
> Are you lookiung for this?
>         echo {{a..z},\ ,{0..9}}{{a..z},\ ,{0..9}}{{a..z},\ ,{0..9}}
>
> I don't want to know why.
> Curiosity killed the cat.
>
> Regards, Mike Jonkmans 


I was going to recommend this:

e=({a..z} ' ' {0..9}); l=${#e[@]}; for ((i=0; i<l; i++)); do for ((j=0; 
j<l; j++)); do for ((y=0; y<l; y++)); do printf "${e[i]}${e[j]}${e[y]} 
"; done; done; done

to make the space complexity, or memory footprint, of it depend linearly 
rather than exponentially (even if just to the 3rd power), on the input, 
allowing more characters in the set to be permuted (e.g. Chinese 
characters) without segfaulting on malloc.

Marco Ippolito
maroloccio@gmail.com