I´ve declared a LOCAL variable (a) inside a function that receives the value of first parameter (${1}). When I call it with a with an asterisk in first parameter´s content, the value assignment had a different treatment. Why?
# function_name () { local a="${1}"; echo $a; }
# set -xv && function_name "param_1_*" && set +xv
set -xv && function_name "param_1_*" && set +xv
+ set -xv
+ function_name 'param_1_*'
+ local 'a=param_1_*'
+ echo 'param_1_*'
param_1_*
+ set +xv
NOTE: + local 'a=param_1_*' <-- See? the quote is inserted before the variable name "a"
# function_name () { a="${1}"; echo $a; }
# set -xv && function_name "param_1_*" && set +xv
+ function_name 'param_1_*'
+ a='param_1_*'
+ echo 'param_1_*'
param_1_*
+ set +xv
NOTE: + a='param_1_*' <-- On this case, the quote is inserted after the equal signal
They're being displayed differently by bash's -x mode, but that doesn't mean bash is treating them differently. What's happening here is that when bash executes local a="${1}", it parses it into some internal representation (something more complicated than a string), substitutes the first parameter into that, and then notices that -x mode is active. So what it does is take that internal representation, and un-parses it into a command you could type in to get the same effect. In general, there will be a number of ways to type in a command that'd produce the same effect (i.e. the internal representation), and it chooses one of these. It could print any of the following:
+ local a='param_1_*'
+ local 'a=param_1_*'
+ local $'a=param_1_*'
+ local a=param_1_"*"
+ local a=param_1_\*
... or any of a number of other possibilities. Its choice may not be the one you'd expect, but it's a perfectly valid way of typing in the "same" command.
BTW, as #Rob said, there is a difference in that a='param_1_*' and 'a=param_1_*' are not equivalent commands, while local a='param_1_*' and local 'a=param_1_*' are equivalent. This is because with the local command, the assignment is essentially a parameter, and quoting around it doesn't matter as much as with a standalone assignment. Thus, when displaying a="${1}", it could print any of:
+ a='param_1_*'
+ a="param_1_*"
+ a=param_1_$'*'
+ a=param_1_\*
but not any of these:
+ 'a=param_1_*'
+ "a=param_1_*"
+ $'a=param_1_*'
I don't think the asterisk is the difference. In your two functions, one is declaring a local and one is setting a shell variable. In the first case you're passing a=param_1_* as an argument to the local builtin (and the shell doesn't care that one of those characters is an * or an =), but in the second the shell is parsing the a=foo into its components. In the first case, it makes sense to put quotation marks around the entire argument, which is a=param_1_*, but it doesn't make sense in the second, since you couldn't really type 'a=param_1_*' at a shell prompt and expect the shell to set the a variable.
Related
Why does this give no output (other than newline) instead of "foo"? The code uses a nameref, which was introduced in bash 4.3, and is a "reference to another variable" which "allows variables to be manipulated indirectly."
And what should be done to guard against this, if writing code for a library?
#!/usr/bin/bash
setret() {
local -n ret_ref=$1
local ret="foo"
ret_ref=$ret
}
setret ret
echo $ret
Running it through bash -x made my head spin, because it looks like it should be outputting the foo that I expected:
+ setret ret
+ local -n ret_ref=ret
+ local ret=foo
+ ret_ref=foo
+ echo
Interestingly, this prints bar, not foo.
#!/usr/bin/bash
setret() {
local -n ret_ref=$1
ret_ref="bar"
local ret="foo"
ret_ref=$ret
}
setret ret
echo $ret
With an equally confusing bash -x output:
+ setret ret
+ local -n ret_ref=ret
+ ret_ref=bar
+ local ret=foo
+ ret_ref=foo
+ echo bar
bar
I'm hoping this is valuable to others, because asking for the expected output in the #bash IRC channel got a response from one of its regulars of foo, which is what I expected.
Then, they set me straight. namerefs just don't work like I thought they did. local -n isn't setting ret_ref to refer to $1. Rather, it basically storing the string ret in ret_ref, marked to be used as a reference when it's used.
So, although it looked to me like ret_ref would refer to the caller's ret variable, it only does so until the function defines its own local ret variable, then it will refer to that one instead.
The only guaranteed way to guard against this, if writing code for a library, is within any function that uses namerefs, to prefix all non-nameref variables with the function name, along these lines:
#!/usr/bin/bash
setret() {
local -n ___setret_ret_ref=$1
local ___setret_ret="foo"
___setret_ret_ref=$___setret_ret
}
setret ret
echo $ret
Very ugly, but necessary to avoid collisions. (Sure, there's less ugly ways to do it that might be likely to work, but not as certain.)
I'm writing a simple graphing script that uses gnuplot, and I use a helper function to construct a .gscript file.
add_gscript() {
echo "->>"
echo $1
echo $1 >> plot.gscript
cat plot.gscript
}
However after passing the following argument into the function
echo "--"
echo 'set xrange [0:$RANGE]'
echo "--"
add_gscript "set xrange [0:100]"
where $RANGE has been defined beforehand, I get the following output
--
set xrange [0:$RANGE]
--
->>
set xrange 1
set datafile separator ","
set term qt size 800,640
set size ratio .618
set xrange 1
Is bash evaluating [0:100] to 1 somehow?
A fix was accurately described in comments on the question:
Always double-quote variable references to have their values treated as literals; without quoting, the values are (in most contexts) subject to shell expansions, including word splitting and pathname expansion. [0:100] happens to be a valid globbing pattern that matches any file in the current dir. named either 0 or : or 1. – mklement0
so echo "$1"; echo "$1" >> plot.gscript and any other unquoted vars. Good luck. – shellter
Double quoting the variables did indeed fix my issue. Thanks!
I'm unable to substitute the argument value(s) in the bash command as below:
# echo $int1
{"id":"74953939-fd20-4472-8aaa-067e6f4c4106"}
# echo $int2
{"id":"5ef4664d-3600-4df9-a6a9-01ffb0f49422"}
# echo $int3
{"id":"6dc95c01-742e-4225-8298-e5750fe67f27"}
# set -x
# data set net-agent interfaces '["$int1", "$int2", "$int3"]'
+ data set net-agent interfaces '["$int1", "$int2", "$int3"]'
Any idea on why the values are not being substituted?
Thanks!
I'm guessing that the argument to the command should be valid JSON, in which case you should remove the double quotes from around each variable and wrap the entire string in double quotes so variables are expanded:
data set net-agent interfaces "[$int1, $int2, $int3]"
Using set -x, this produces:
$ data set net-agent interfaces "[$int1, $int2, $int3]"
+ data set net-agent interfaces '[{"id":"74953939-fd20-4472-8aaa-067e6f4c4106"}, {"id":"5ef4664d-3600-4df9-a6a9-01ffb0f49422"}, {"id":"6dc95c01-742e-4225-8298-e5750fe67f27"}]'
Consider this PS1
PS1='\n${_:+$? }$ '
Here is the result of a few commands
$ [ 2 = 2 ]
0 $ [ 2 = 3 ]
1 $
1 $
The first line shows no status as expected, and the next two lines show the
correct exit code. However on line 3 only Enter was pressed, so I would like the
status to go away, like line 1. How can I do this?
Here's a funny, very simple possibility: it uses the \# escape sequence of PS1 together with parameter expansions (and the way Bash expands its prompt).
The escape sequence \# expands to the command number of the command to be executed. This is incremented each time a command has actually been executed. Try it:
$ PS1='\# $ '
2 $ echo hello
hello
3 $ # this is a comment
3 $
3 $ echo hello
hello
4 $
Now, each time a prompt is to be displayed, Bash first expands the escape sequences found in PS1, then (provided the shell option promptvars is set, which is the default), this string is expanded via parameter expansion, command substitution, arithmetic expansion, and quote removal.
The trick is then to have an array that will have the k-th field set (to the empty string) whenever the (k-1)-th command is executed. Then, using appropriate parameter expansions, we'll be able to detect when these fields are set and to display the return code of the previous command if the field isn't set. If you want to call this array __cmdnbary, just do:
PS1='\n${__cmdnbary[\#]-$? }${__cmdnbary[\#]=}\$ '
Look:
$ PS1='\n${__cmdnbary[\#]-$? }${__cmdnbary[\#]=}\$ '
0 $ [ 2 = 3 ]
1 $
$ # it seems that it works
$ echo "it works"
it works
0 $
To qualify for the shortest answer challenge:
PS1='\n${a[\#]-$? }${a[\#]=}$ '
that's 31 characters.
Don't use this, of course, as a is a too trivial name; also, \$ might be better than $.
Seems you don't like that the initial prompt is 0 $; you can very easily modify this by initializing the array __cmdnbary appropriately: you'll put this somewhere in your configuration file:
__cmdnbary=( '' '' ) # Initialize the field 1!
PS1='\n${__cmdnbary[\#]-$? }${__cmdnbary[\#]=}\$ '
Got some time to play around this weekend. Looking at my earlier answer (not-good) and other answers I think this may be probably the smallest answer.
Place these lines at the end of your ~/.bash_profile:
PS1='$_ret$ '
trapDbg() {
local c="$BASH_COMMAND"
[[ "$c" != "pc" ]] && export _cmd="$c"
}
pc() {
local r=$?
trap "" DEBUG
[[ -n "$_cmd" ]] && _ret="$r " || _ret=""
export _ret
export _cmd=
trap 'trapDbg' DEBUG
}
export PROMPT_COMMAND=pc
trap 'trapDbg' DEBUG
Then open a new terminal and note this desired behavior on BASH prompt:
$ uname
Darwin
0 $
$
$
$ date
Sun Dec 14 05:59:03 EST 2014
0 $
$
$ [ 1 = 2 ]
1 $
$
$ ls 123
ls: cannot access 123: No such file or directory
2 $
$
Explanation:
This is based on trap 'handler' DEBUG and PROMPT_COMMAND hooks.
PS1 is using a variable _ret i.e. PS1='$_ret$ '.
trap command runs only when a command is executed but PROMPT_COMMAND is run even when an empty enter is pressed.
trap command sets a variable _cmd to the actually executed command using BASH internal var BASH_COMMAND.
PROMPT_COMMAND hook sets _ret to "$? " if _cmd is non-empty otherwise sets _ret to "". Finally it resets _cmd var to empty state.
The variable HISTCMD is updated every time a new command is executed. Unfortunately, the value is masked during the execution of PROMPT_COMMAND (I suppose for reasons related to not having history messed up with things which happen in the prompt command). The workaround I came up with is kind of messy, but it seems to work in my limited testing.
# This only works if the prompt has a prefix
# which is displayed before the status code field.
# Fortunately, in this case, there is one.
# Maybe use a no-op prefix in the worst case (!)
PS1_base=$'\n'
# Functions for PROMPT_COMMAND
PS1_update_HISTCMD () {
# If HISTCONTROL contains "ignoredups" or "ignoreboth", this breaks.
# We should not change it programmatically
# (think principle of least astonishment etc)
# but we can always gripe.
case :$HISTCONTROL: in
*:ignoredups:* | *:ignoreboth:* )
echo "PS1_update_HISTCMD(): HISTCONTROL contains 'ignoredups' or 'ignoreboth'" >&2
echo "PS1_update_HISTCMD(): Warning: Please remove this setting." >&2 ;;
esac
# PS1_HISTCMD needs to contain the old value of PS1_HISTCMD2 (a copy of HISTCMD)
PS1_HISTCMD=${PS1_HISTCMD2:-$PS1_HISTCMD}
# PS1_HISTCMD2 needs to be unset for the next prompt to trigger properly
unset PS1_HISTCMD2
}
PROMPT_COMMAND=PS1_update_HISTCMD
# Finally, the actual prompt:
PS1='${PS1_base#foo${PS1_HISTCMD2:=${HISTCMD%$PS1_HISTCMD}}}${_:+${PS1_HISTCMD2:+$? }}$ '
The logic in the prompt is roughly as follows:
${PS1_base#foo...}
This displays the prefix. The stuff in #... is useful only for its side effects. We want to do some variable manipulation without having the values of the variables display, so we hide them in a string substitution. (This will display odd and possibly spectacular things if the value of PS1_base ever happens to begin with foo followed by the current command history index.)
${PS1_HISTCMD2:=...}
This assigns a value to PS1_HISTCMD2 (if it is unset, which we have made sure it is). The substitution would nominally also expand to the new value, but we have hidden it in a ${var#subst} as explained above.
${HISTCMD%$PS1_HISTCMD}
We assign either the value of HISTCMD (when a new entry in the command history is being made, i.e. we are executing a new command) or an empty string (when the command is empty) to PS1_HISTCMD2. This works by trimming off the value HISTCMD any match on PS1_HISTCMD (using the ${var%subst} suffix replacement syntax).
${_:+...}
This is from the question. It will expand to ... something if the value of $_ is set and nonempty (which it is when a command is being executed, but not e.g. if we are performing a variable assignment). The "something" should be the status code (and a space, for legibility) if PS1_HISTCMD2 is nonempty.
${PS1_HISTCMD2:+$? }
There.
'$ '
This is just the actual prompt suffix, as in the original question.
So the key parts are the variables PS1_HISTCMD which remembers the previous value of HISTCMD, and the variable PS1_HISTCMD2 which captures the value of HISTCMD so it can be accessed from within PROMPT_COMMAND, but needs to be unset in the PROMPT_COMMAND so that the ${PS1_HISTCMD2:=...} assignment will fire again the next time the prompt is displayed.
I fiddled for a bit with trying to hide the output from ${PS1_HISTCMD2:=...} but then realized that there is in fact something we want to display anyhow, so just piggyback on that. You can't have a completely empty PS1_base because the shell apparently notices, and does not even attempt to perform a substitution when there is no value; but perhaps you can come up with a dummy value (a no-op escape sequence, perhaps?) if you have nothing else you want to display. Or maybe this could be refactored to run with a suffix instead; but that is probably going to be trickier still.
In response to Anubhava's "smallest answer" challenge, here is the code without comments or error checking.
PS1_base=$'\n'
PS1_update_HISTCMD () { PS1_HISTCMD=${PS1_HISTCMD2:-$PS1_HISTCMD}; unset PS1_HISTCMD2; }
PROMPT_COMMAND=PS1_update_HISTCMD
PS1='${PS1_base#foo${PS1_HISTCMD2:=${HISTCMD%$PS1_HISTCMD}}}${_:+${PS1_HISTCMD2:+$? }}$ '
This is probably not the best way to do this, but it seems to be working
function pc {
foo=$_
fc -l > /tmp/new
if cmp -s /tmp/{new,old} || test -z "$foo"
then
PS1='\n$ '
else
PS1='\n$? $ '
fi
cp /tmp/{new,old}
}
PROMPT_COMMAND=pc
Result
$ [ 2 = 2 ]
0 $ [ 2 = 3 ]
1 $
$
I need to use great script bash-preexec.sh.
Although I don't like external dependencies, this was the only thing to help me avoid to have 1 in $? after just pressing enter without running any command.
This goes to your ~/.bashrc:
__prompt_command() {
local exit="$?"
PS1='\u#\h: \w \$ '
[ -n "$LASTCMD" -a "$exit" != "0" ] && PS1='['${red}$exit$clear"] $PS1"
}
PROMPT_COMMAND=__prompt_command
[-f ~/.bash-preexec.sh ] && . ~/.bash-preexec.sh
preexec() { LASTCMD="$1"; }
UPDATE: later I was able to find a solution without dependency on .bash-preexec.sh.
I am changing a bash script which has a structure as such:
#somewhere in the code
sim_counts=#... some value
function_name()
{
set $sim_counts
for hostname in $linux_hostnames; do
if [ $1 -eq 0 ]; then # if sim_counts equal 0
shift # jump forward in sim_counts
continue
fi
# ... more code
shift
done
}
Then it is called in the script:
function_name
I want to introduce a parameter to this function:
#somewhere in the code
sim_counts=#... some value
function_name()
{
ip=$1
set $sim_counts
for hostname in $linux_hostnames; do
if [ $1 -eq 0 ]; then # if sim_counts equal 0
shift # jump forward in sim_counts
continue
fi
# ... more code
shift
done
}
And call the function in following way:
function_name 10.255.192.123
What should I do to avoid $1 conflict of function parameter and the other value from set command ?
If I am correctly reading the set builtin page in the Bash Reference Manual, I believe the code as you have written it will just work. Quoting from that page:
The remaining N arguments are positional parameters and are assigned, in order, to $1, $2, … $N. The special parameter # is set to N.
In essence, any pre-existing values for the positional variables will be blown away. The first sentence on that manual page is also interesting:
This builtin is so complicated that it deserves its own section.
In short, I think your code should simply work as expected. You've saved the initial value of $1 (from the function call) into a temporary variable; as long as you refer to $ip for that specific value, you should be good. In my own test script, it seems that $1 gets blown away as I expect it should.