If I ask bash to echo the -1th argument it prints hb1:
echo $-1
hb1
Why? What is it accessing?
5
20
You are not asking it to print the 1st argument, that would be: $1.
What you are asking for is a special parameter:
-(
$-, a hyphen.) Expands to the current option flags as specified upon invocation, by the set builtin command, or those set by the shell itself (such as the -i option).
So your options are: hb
Then you see the 1 you've added is printed afterwards (hence hb1).
If you are looking to get the last argument (Not sure if that is what you meant by -1 argument), you can use Shell Parameter Expansion in the following form:
$ set -- one two three
$ echo "${@: -1}"
three
-
8
${!#}is also the value of the last parameter, using indirect expansion. – glenn jackman Jun 22 at 19:30 -
2
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@Isaac If you're lucky, that's true. If
IFScontains any digits in the value$#expands to, not so much. Always safer to quote;eval "echo \"\$$#\""is more reliable. To see the failure for yourself:IFS=1234567890; set -- one two three; eval echo \$$#only outputs$. – Charles Duffy Jun 23 at 20:10 -
@CharlesDuffy Yes, you are correct, my mistake. But it is also required to allow positional parameters at position 10 and up. For a really reliable solution add {} :
IFS=1234567890; set -- {a..w}; eval echo \"\$\{"$#"\}\";-) – Isaac Jun 23 at 20:27
$-gets the current shell options (which is apparently "hb"). What did you expect it to do? – Gordon Davisson Jun 22 at 19:21