On Sat, Feb 5, 2022 at 9:39 PM Alex fxmbsw7 Ratchev <fxmb...@gmail.com> wrote:
>
> On Sat, Feb 5, 2022 at 7:55 PM Chet Ramey <chet.ra...@case.edu> wrote:
> >
> > On 2/4/22 6:17 PM, Alex fxmbsw7 Ratchev wrote:
> > > what about this viewing point
> > > aliases can start, $(('s, but not end... this is unlogic
> >
> > Well, I don't know about `unlogic' but there's an understanding deficit,
> > for sure.
> >
> > >
> > > alias -- \
> > > p='printf %s\\n ' \
> > > assign='assign=$(( ' begin='$(( ' \
> > >
> > > for data in "1 + 2"
> > > do
> > > alias -- data="$d "
> > > p begin data ))
> > > done
> > >
> > > this works and results 3
> >
> > OK, let's go through it. I'll gloss over some non-essential details and
> > simplify others. I'll skip over the bulk of the for loop parsing and just
> > cover the simple command where alias expansion takes place.
> >
> > Since `p' is in a command position, it gets alias expanded to
> > `printf %s\\n '. The lexical analysis restarts, the two words get scanned,
> > and we start a simple command. The expansion ends with a space, so the next
> > token (`begin') undergoes alias expansion, gets expanded to `$(( ', and the
> > lex restarts. We read `$((' and know we're reading the start of a word that
> > begins with an arithmetic expansion. We keep reading as if we are reading a
> > double-quoted string, looking for `))', since that's what you do while
> > reading an arithmetic expansion.
>
> so the alias expansion fails at the current command is no alias
> anymore, so alias parsing rules dont apply anymore, .. no ... hmm .. ?
> would make much sense then and i have nothing else to say..maybe ask about an shopt feature to enable straight alias expansion when the aliases are following .. cheers and peace > btw the printf leet were examples about the other, not to work, not to > say i tried to code so > > > There is no alias expansion because we are > > not reading a new token and the contents of the arithmetic expansion are > > treated like a double-quoted string, both disqualifying conditions. We find > > those two characters, closing the arithmetic expansion, and go on looking > > for the end of the word. We read a newline, which delimits the token, which > > is the (shell grammar) WORD `$(( data ))'. > > > > After we finish parsing the for loop, we go back to execute it. We run > > a useless `alias' command, then come to the simple command > > > > printf %s\\n $((data)) > > > > (I removed the spaces from the arithmetic expansion to make it clear that > > it's one word.) > > > > We begin to expand the words in this simple command. When we come to the > > arithmetic expansion, we take the contents (`data') and run them through > > the arithmetic evaluator. We get an identifier (`data'), assume it's a > > shell variable, look up its value (`1 + 2', since that's how the for loop > > set it) and evaluate that as an expression. It evaluates to 3, and the > > string "3" becomes the result of the arithmetic expansion. > > > > We run > > > > printf %s\n 3 > > > > and get `3' as output. > > > > Understand? > > yes, now, fail at no more alias so no alias expanded > > cheers, thank you much for your kind good effords > > > -- > > ``The lyf so short, the craft so long to lerne.'' - Chaucer > > ``Ars longa, vita brevis'' - Hippocrates > > Chet Ramey, UTech, CWRU c...@case.edu http://tiswww.cwru.edu/~chet/
