Re: Counting characters in command output?

[Ray Andrews <rayandrews@eastlink.ca>]

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On 2024-02-14 07:58, Mark J. Reed wrote:
> > I myself almost always want lines.  Or elements.  Sometimes words, 
> almost  never characters.
>
> As I said, it's the consistency.
Right.  That would be second on my list of vectors and should trump 
tradition, especially when, as Roman says, most users don't prefer the 
status quo.

> If you have a scalar parameter named *foo*, *$#foo* is the number of 
> characters. If it's an array, *$#foo* is the number of elements.  The 
> key point in my mind is that when you assign a parameter from command 
> substitution, e.g. *foo=$(bar)*, then what you get is a scalar 
> parameter, not an array. You /can/ get an array instead, but you have 
> to ask for it explicitly by putting extra parentheses around the right 
> hand side: *foo=($(bar))*. So parameter assignment defaults to scalar 
> mode and requires extra punctuation to do array mode. But directly 
> counting the result of the substitution with no intervening parameter 
> defaults to array mode and requires extra punctuation to do scalar mode.
So really there is a logical problem too.  We have another 'invisible' 
transformation.  We 'have' quarts but liters are what's counted, yes?

 > I would in general have expected *foo $(bar) * to behave 
identically to *baz=$(bar); foo $baz*.

Yes, it would be non-negotiable in algebra.  a=b; b=c; ergo a=c.

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