Chet Ramey <chet.ra...@case.edu> writes:
> Bash allows the close brace to be joined to the remaining
> characters in the word without being followed by a shell metacharacter
> as a reserved word would usually require.
I had to read this a couple of times to figure out what it means. In
particular "the word" isn't well-bound here. Perhaps better is
> The characters immediately following the close brace continue the word
> that the command substitution is part of; the close brace need not be
> followed by a shell metacharacter as a reserved word would usually
> require.
This text is clear in its description of the behavior but entirely
unclear regarding what the point is:
> If the first character is a '|', the construct expands to the value of
> the 'REPLY' shell variable after COMMAND executes, without removing any
> trailing newlines, and the standard output of COMMAND remains the same
> as in the calling shell. Bash creates 'REPLY' as an initially-unset
> local variable when COMMAND executes, and restores 'REPLY' to the value
> it had before the command substitution after COMMAND completes, as with
> any local variable.
My guess is that the intention is for COMMAND to be "read" with no
names:
${| read; }
If so, can you make this an example for the "${|" form?
Also, can you clarify the semantics on this point: The text above says
"the construct expands to the value of the 'REPLY' shell variable after
COMMAND executes, without removing any trailing newlines". However, the
documentation of "read" says: "If no names are supplied, the line read,
without the ending delimiter but otherwise unmodified, is assigned to
the variable REPLY."
So the sequence of processing is: read inputs a line; read deletes the
ending newline and assigns the remainder to REPLY; command substitution
obtains the value of REPLY (which does not contain a newline); command
substitution returns that value "without removing any trailing
newlines".
Is that what you intend the code to do/what the code does?
Dale
