Linux

Code:

aa=|
echo $aa

The above echo works but the below echo fails. Why please?

Code:

IFS=:
aa=|
echo $aa
echo $IFS

The later 'echo' command will work if variable is put in codes.

Code:

echo "$aa"
echo "$IFS"

I summarize that when IFS is set to ':' or '|', echo used with variable doesn't work unless the variable is quoted.
It took quite some time to conclude this.

All string arguments, regardless if literal or variables, should be quoted:

Code:

echo "$aa"
echo "$IFS"

Special characters in a literal assignment should be quoted, too:

Code:

aa="|"

Does it? I don't think so. The | char opens a pipe and makes the shell wait for the next command in the pipe.

The shell command parsing step which looks for pipes has already concluded by the time parameter expansion occurs. There's nothing special about a pipe character in a variable. You'd have to use eval to make it so.

---------- Post updated at 10:10 AM ---------- Previous update was at 10:01 AM ----------

What's happening is perfectly normal. It's called field splitting.

Field splitting is one of the final steps in sh command parsing. After the variable is replaced with its value, that value is split into fields using IFS. An IFS character will never be seen in a command argument unless it's quoted (since quoting tells the shell not to perform field splitting).

Regards,
Alister

The command that was given in post#1 was

Code:

aa=|
echo $aa

Executing this on the two systems I have at hand (Linux, FreeBSD) issues the secondary prompt (asking for the next command in pipe, I assume). The variable aa is not assigned to, it is not even defined afterwards. So I think my statement holds.
Of course, assigning an escaped | is totally different and absolutely OK.

It does indeed. My mistake.

What I said is technically correct, but it does not apply here since the pipe is unquoted and never assigned to the variable.

Regards,
Alister

Yes Rudic, you are right. One mistake from my side:
In place of "|" , it should be ":".

Code:

IFS=:
aa=:
echo $aa

Now echo won't say the output because of the reason given by Alister.
Thanks Alister for the elaborate clarification.