$_ vs !$. Last argument of the preceding command and output redirection

Question

The question is about special variables. Documentation says:

!!:$

designates the last argument of the preceding command. This may be shortened to !$.

---

($_, an underscore.) At shell startup, set to the absolute pathname used to invoke the shell or shell script being executed as passed in the environment or argument list. Subsequently, expands to the last argument to the previous command after expansion. Also set to the full pathname used to invoke each command executed and placed in the environment exported to that command.

There must be some difference I cannot catch, because:

$ echo "hello" > /tmp/a.txt
$ echo "!$"
echo "/tmp/a.txt"
/tmp/a.txt

$ echo "hello" > /tmp/a.txt
$ echo $_
hello

What is the difference?

Answer 1

!$ is a word designator of history expansion; it expands to the last word of the previous command in history. In other words, the last word of the previous entry in history. This word is usually the last argument to the command, but not in case of redirection. In:

echo "hello" > /tmp/a.txt

the whole command 'echo "hello" > /tmp/a.txt' appeared in history, and /tmp/a.txt is the last word of that command.

_ is a shell parameter; it expands to the last argument of the previous command. Here, the redirection is not a part of arguments passed to the command, so only hello is the argument passed to echo. That's why $_ expanded to hello.

_ is no longer one of shell standard special parameters. It works in bash, zsh, mksh and dash only when interactive, ksh93 only when two commands are on separated lines:

$ echo 1 && echo $_
1
/usr/bin/ksh

$ echo 1
1
$ echo $_
1

Answer 2

I found it difficult if not impossible to use parameter substitution directly with !$.

For example if I could wget http://foo.com/my.tar then tar xf ${_##*/} but with !$ it is not possible without defining an intermediary variable.