Re: Equivalent of Perl 'join' ?

[CdeMills]

John W. Eaton wrote:
> 
> 
> or
> 
>   join = @(expr, list) [(sprintf (["%s", expr], list{1:end-1})),
> list{end}]
> 
> Have we beaten this topic to death yet?
> 
> 

Your solution is indeed clean and  elegant. But the first poster who came
with the idea of using sprintf introduced a sensitivity to 'expr' value:
join('%', {'a', 'b', 'c'}) 
will fail. I know the right form is 
join('%%', , {'a', 'b', 'c'}) , 
but still this is an interference from the internal 'engine'. Now, this
won't prevent me of sleeping well :-)

Regards 

Pascal
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