Bash how to print out all the input arguments without using loop - bash

I've heard of that loop in bash is in efficient and complicated. so i'm try to make a script that print out all the argument in a specific format without using any loop.
When i do
bash script_name arg1 arg2 arg3...
or
./script_name arg1 arg2 arg3...
it should output:
0: script_name
1: arg1
2: arg2
...
I try to use $# but the problem is it never prints out the script name, and i'm having trouble to come up with a way to print out the index.
can anyone pls give me a hint? thank you

Firstly, $# does not contain the script name, it only contains the actual command-line arguments. You have to print $0 explicitly to print the script name.
Here are two solutions:
exec paste -d: <(seq 0 $#;) <(printf ' %s\n' "$0" "$#";);
The above runs the paste utility to paste together (1) a sequence of numbers from 0 to the number of arguments given, generated by the seq utility, and (2) the script name and command-line arguments, printed by printf, one per line. Each of the seq and printf commands is run in a process substitution construct, with each generated device file representing the command output (e.g. /dev/fd/63 on my system) passed to paste.
exec printf '%s\n' "$0" "$#"| awk '{ print(i++": "$0); };';
The above prints the script name and command-line arguments, one per line, and pipes them to awk, which adds the desired numbering prefix.
It should be noted that each of these two solutions involves starting at least one additional system process, either for process substitution or for a pipeline. That is pretty much guaranteed to be much more inefficient than running the bash built-in for-loop, which requires no additional processes (assuming that you only call built-in stuff within the loop, e.g. printf). In general, built-in commands/keywords/constructs will almost always be faster than running external executables.

Related

What shellenv command does? [duplicate]

After reading the Bash man pages and with respect to this post, I am still having trouble understanding what exactly the eval command does and which would be its typical uses.
For example, if we do:
$ set -- one two three # Sets $1 $2 $3
$ echo $1
one
$ n=1
$ echo ${$n} ## First attempt to echo $1 using brackets fails
bash: ${$n}: bad substitution
$ echo $($n) ## Second attempt to echo $1 using parentheses fails
bash: 1: command not found
$ eval echo \${$n} ## Third attempt to echo $1 using 'eval' succeeds
one
What exactly is happening here and how do the dollar sign and the backslash tie into the problem?
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line. (If you pass several arguments, they are first joined with spaces between them.)
${$n} is a syntax error in bash. Inside the braces, you can only have a variable name, with some possible prefix and suffixes, but you can't have arbitrary bash syntax and in particular you can't use variable expansion. There is a way of saying “the value of the variable whose name is in this variable”, though:
echo ${!n}
one
$(…) runs the command specified inside the parentheses in a subshell (i.e. in a separate process that inherits all settings such as variable values from the current shell), and gathers its output. So echo $($n) runs $n as a shell command, and displays its output. Since $n evaluates to 1, $($n) attempts to run the command 1, which does not exist.
eval echo \${$n} runs the parameters passed to eval. After expansion, the parameters are echo and ${1}. So eval echo \${$n} runs the command echo ${1}.
Note that most of the time, you must use double quotes around variable substitutions and command substitutions (i.e. anytime there's a $): "$foo", "$(foo)". Always put double quotes around variable and command substitutions, unless you know you need to leave them off. Without the double quotes, the shell performs field splitting (i.e. it splits value of the variable or the output from the command into separate words) and then treats each word as a wildcard pattern. For example:
$ ls
file1 file2 otherfile
$ set -- 'f* *'
$ echo "$1"
f* *
$ echo $1
file1 file2 file1 file2 otherfile
$ n=1
$ eval echo \${$n}
file1 file2 file1 file2 otherfile
$eval echo \"\${$n}\"
f* *
$ echo "${!n}"
f* *
eval is not used very often. In some shells, the most common use is to obtain the value of a variable whose name is not known until runtime. In bash, this is not necessary thanks to the ${!VAR} syntax. eval is still useful when you need to construct a longer command containing operators, reserved words, etc.
Simply think of eval as "evaluating your expression one additional time before execution"
eval echo \${$n} becomes echo $1 after the first round of evaluation. Three changes to notice:
The \$ became $ (The backslash is needed, otherwise it tries to evaluate ${$n}, which means a variable named {$n}, which is not allowed)
$n was evaluated to 1
The eval disappeared
In the second round, it is basically echo $1 which can be directly executed.
So eval <some command> will first evaluate <some command> (by evaluate here I mean substitute variables, replace escaped characters with the correct ones etc.), and then run the resultant expression once again.
eval is used when you want to dynamically create variables, or to read outputs from programs specifically designed to be read like this. See Eval command and security issues for examples. The link also contains some typical ways in which eval is used, and the risks associated with it.
In my experience, a "typical" use of eval is for running commands that generate shell commands to set environment variables.
Perhaps you have a system that uses a collection of environment variables, and you have a script or program that determines which ones should be set and their values. Whenever you run a script or program, it runs in a forked process, so anything it does directly to environment variables is lost when it exits. But that script or program can send the export commands to standard output.
Without eval, you would need to redirect standard output to a temporary file, source the temporary file, and then delete it. With eval, you can just:
eval "$(script-or-program)"
Note the quotes are important. Take this (contrived) example:
# activate.sh
echo 'I got activated!'
# test.py
print("export foo=bar/baz/womp")
print(". activate.sh")
$ eval $(python test.py)
bash: export: `.': not a valid identifier
bash: export: `activate.sh': not a valid identifier
$ eval "$(python test.py)"
I got activated!
The eval statement tells the shell to take eval’s arguments as commands and run them through the command-line. It is useful in a situation like below:
In your script if you are defining a command into a variable and later on you want to use that command then you should use eval:
a="ls | more"
$a
Output:
bash: command not found: ls | more
The above command didn't work as ls tried to list file with name pipe (|) and more. But these files are not there:
eval $a
Output:
file.txt
mailids
remote_cmd.sh
sample.txt
tmp
Update: Some people say one should -never- use eval. I disagree. I think the risk arises when corrupt input can be passed to eval. However there are many common situations where that is not a risk, and therefore it is worth knowing how to use eval in any case. This stackoverflow answer explains the risks of eval and alternatives to eval. Ultimately it is up to the user to determine if/when eval is safe and efficient to use.
The bash eval statement allows you to execute lines of code calculated or acquired, by your bash script.
Perhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order. That's essentially the same behavior as the bash source statement, which is what one would use, unless it was necessary to perform some kind of transformation (e.g. filtering or substitution) on the content of the imported script.
I rarely have needed eval, but I have found it useful to read or write variables whose names were contained in strings assigned to other variables. For example, to perform actions on sets of variables, while keeping the code footprint small and avoiding redundancy.
eval is conceptually simple. However, the strict syntax of the bash language, and the bash interpreter's parsing order can be nuanced and make eval appear cryptic and difficult to use or understand. Here are the essentials:
The argument passed to eval is a string expression that is calculated at runtime. eval will execute the final parsed result of its argument as an actual line of code in your script.
Syntax and parsing order are stringent. If the result isn't an executable line of bash code, in scope of your script, the program will crash on the eval statement as it tries to execute garbage.
When testing you can replace the eval statement with echo and look at what is displayed. If it is legitimate code in the current context, running it through eval will work.
The following examples may help clarify how eval works...
Example 1:
eval statement in front of 'normal' code is a NOP
$ eval a=b
$ eval echo $a
b
In the above example, the first eval statements has no purpose and can be eliminated. eval is pointless in the first line because there is no dynamic aspect to the code, i.e. it already parsed into the final lines of bash code, thus it would be identical as a normal statement of code in the bash script. The 2nd eval is pointless too, because, although there is a parsing step converting $a to its literal string equivalent, there is no indirection (e.g. no referencing via string value of an actual bash noun or bash-held script variable), so it would behave identically as a line of code without the eval prefix.
Example 2:
Perform var assignment using var names passed as string values.
$ key="mykey"
$ val="myval"
$ eval $key=$val
$ echo $mykey
myval
If you were to echo $key=$val, the output would be:
mykey=myval
That, being the final result of string parsing, is what will be executed by eval, hence the result of the echo statement at the end...
Example 3:
Adding more indirection to Example 2
$ keyA="keyB"
$ valA="valB"
$ keyB="that"
$ valB="amazing"
$ eval eval \$$keyA=\$$valA
$ echo $that
amazing
The above is a bit more complicated than the previous example, relying more heavily on the parsing-order and peculiarities of bash. The eval line would roughly get parsed internally in the following order (note the following statements are pseudocode, not real code, just to attempt to show how the statement would get broken down into steps internally to arrive at the final result).
eval eval \$$keyA=\$$valA # substitution of $keyA and $valA by interpreter
eval eval \$keyB=\$valB # convert '$' + name-strings to real vars by eval
eval $keyB=$valB # substitution of $keyB and $valB by interpreter
eval that=amazing # execute string literal 'that=amazing' by eval
If the assumed parsing order doesn't explain what eval is doing enough, the third example may describe the parsing in more detail to help clarify what is going on.
Example 4:
Discover whether vars, whose names are contained in strings, themselves contain string values.
a="User-provided"
b="Another user-provided optional value"
c=""
myvarname_a="a"
myvarname_b="b"
myvarname_c="c"
for varname in "myvarname_a" "myvarname_b" "myvarname_c"; do
eval varval=\$$varname
if [ -z "$varval" ]; then
read -p "$varname? " $varname
fi
done
In the first iteration:
varname="myvarname_a"
Bash parses the argument to eval, and eval sees literally this at runtime:
eval varval=\$$myvarname_a
The following pseudocode attempts to illustrate how bash interprets the above line of real code, to arrive at the final value executed by eval. (the following lines descriptive, not exact bash code):
1. eval varval="\$" + "$varname" # This substitution resolved in eval statement
2. .................. "$myvarname_a" # $myvarname_a previously resolved by for-loop
3. .................. "a" # ... to this value
4. eval "varval=$a" # This requires one more parsing step
5. eval varval="User-provided" # Final result of parsing (eval executes this)
Once all the parsing is done, the result is what is executed, and its effect is obvious, demonstrating there is nothing particularly mysterious about eval itself, and the complexity is in the parsing of its argument.
varval="User-provided"
The remaining code in the example above simply tests to see if the value assigned to $varval is null, and, if so, prompts the user to provide a value.
I originally intentionally never learned how to use eval, because most people will recommend to stay away from it like the plague. However I recently discovered a use case that made me facepalm for not recognizing it sooner.
If you have cron jobs that you want to run interactively to test, you might view the contents of the file with cat, and copy and paste the cron job to run it. Unfortunately, this involves touching the mouse, which is a sin in my book.
Lets say you have a cron job at /etc/cron.d/repeatme with the contents:
*/10 * * * * root program arg1 arg2
You cant execute this as a script with all the junk in front of it, but we can use cut to get rid of all the junk, wrap it in a subshell, and execute the string with eval
eval $( cut -d ' ' -f 6- /etc/cron.d/repeatme)
The cut command only prints out the 6th field of the file, delimited by spaces. Eval then executes that command.
I used a cron job here as an example, but the concept is to format text from stdout, and then evaluate that text.
The use of eval in this case is not insecure, because we know exactly what we will be evaluating before hand.
I've recently had to use eval to force multiple brace expansions to be evaluated in the order I needed. Bash does multiple brace expansions from left to right, so
xargs -I_ cat _/{11..15}/{8..5}.jpg
expands to
xargs -I_ cat _/11/8.jpg _/11/7.jpg _/11/6.jpg _/11/5.jpg _/12/8.jpg _/12/7.jpg _/12/6.jpg _/12/5.jpg _/13/8.jpg _/13/7.jpg _/13/6.jpg _/13/5.jpg _/14/8.jpg _/14/7.jpg _/14/6.jpg _/14/5.jpg _/15/8.jpg _/15/7.jpg _/15/6.jpg _/15/5.jpg
but I needed the second brace expansion done first, yielding
xargs -I_ cat _/11/8.jpg _/12/8.jpg _/13/8.jpg _/14/8.jpg _/15/8.jpg _/11/7.jpg _/12/7.jpg _/13/7.jpg _/14/7.jpg _/15/7.jpg _/11/6.jpg _/12/6.jpg _/13/6.jpg _/14/6.jpg _/15/6.jpg _/11/5.jpg _/12/5.jpg _/13/5.jpg _/14/5.jpg _/15/5.jpg
The best I could come up with to do that was
xargs -I_ cat $(eval echo _/'{11..15}'/{8..5}.jpg)
This works because the single quotes protect the first set of braces from expansion during the parsing of the eval command line, leaving them to be expanded by the subshell invoked by eval.
There may be some cunning scheme involving nested brace expansions that allows this to happen in one step, but if there is I'm too old and stupid to see it.
You asked about typical uses.
One common complaint about shell scripting is that you (allegedly) can't pass by reference to get values back out of functions.
But actually, via "eval", you can pass by reference. The callee can pass back a list of variable assignments to be evaluated by the caller. It is pass by reference because the caller can allowed to specify the name(s) of the result variable(s) - see example below. Error results can be passed back standard names like errno and errstr.
Here is an example of passing by reference in bash:
#!/bin/bash
isint()
{
re='^[-]?[0-9]+$'
[[ $1 =~ $re ]]
}
#args 1: name of result variable, 2: first addend, 3: second addend
iadd()
{
if isint ${2} && isint ${3} ; then
echo "$1=$((${2}+${3}));errno=0"
return 0
else
echo "errstr=\"Error: non-integer argument to iadd $*\" ; errno=329"
return 1
fi
}
var=1
echo "[1] var=$var"
eval $(iadd var A B)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[2] var=$var (unchanged after error)"
eval $(iadd var $var 1)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[3] var=$var (successfully changed)"
The output looks like this:
[1] var=1
errstr=Error: non-integer argument to iadd var A B
errno=329
[2] var=1 (unchanged after error)
[3] var=2 (successfully changed)
There is almost unlimited band width in that text output! And there are more possibilities if the multiple output lines are used: e.g., the first line could be used for variable assignments, the second for continuous 'stream of thought', but that's beyond the scope of this post.
In the question:
who | grep $(tty | sed s:/dev/::)
outputs errors claiming that files a and tty do not exist. I understood this to mean that tty is not being interpreted before execution of grep, but instead that bash passed tty as a parameter to grep, which interpreted it as a file name.
There is also a situation of nested redirection, which should be handled by matched parentheses which should specify a child process, but bash is primitively a word separator, creating parameters to be sent to a program, therefore parentheses are not matched first, but interpreted as seen.
I got specific with grep, and specified the file as a parameter instead of using a pipe. I also simplified the base command, passing output from a command as a file, so that i/o piping would not be nested:
grep $(tty | sed s:/dev/::) <(who)
works well.
who | grep $(echo pts/3)
is not really desired, but eliminates the nested pipe and also works well.
In conclusion, bash does not seem to like nested pipping. It is important to understand that bash is not a new-wave program written in a recursive manner. Instead, bash is an old 1,2,3 program, which has been appended with features. For purposes of assuring backward compatibility, the initial manner of interpretation has never been modified. If bash was rewritten to first match parentheses, how many bugs would be introduced into how many bash programs? Many programmers love to be cryptic.
As clearlight has said, "(p)erhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order". I'm no expert, but the textbook I'm currently reading (Shell-Programmierung by Jürgen Wolf) points to one particular use of this that I think would be a valuable addition to the set of potential use cases collected here.
For debugging purposes, you may want to go through your script line by line (pressing Enter for each step). You could use eval to execute every line by trapping the DEBUG signal (which I think is sent after every line):
trap 'printf "$LINENO :-> " ; read line ; eval $line' DEBUG
I like the "evaluating your expression one additional time before execution" answer, and would like to clarify with another example.
var="\"par1 par2\""
echo $var # prints nicely "par1 par2"
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "par1 par2" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
# Option 1: Will Pass
echo "eval \"cntpars \$var\""
eval "cntpars $var"
# Option 2: Will Fail, with curious results
echo "cntpars \$var"
cntpars $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?

bash: filenames as parameter, perform action in cycle

My current script goes like:
#!/bin/bash
victims=*asci*
for f in $victims ; do
awk /some blah blah here/ ;done
so basically takes all files containing ascii in their name and performs an action on them.
I wanted, however, the filenames be entered as a parameter. Like:
bash myscript.sh *log* for example.
When using
#!/bin/bash
victims="$1"
for f in $victims ; do
awk /some blah blah here/ ;done
it doesnt do what expected. Performs only on the first file (as far as I remember).
May I ask for a help? Want the script to perform a function over a bunch of files that contain the parameter in their filename. Im not very experienced in bash, honestly. Thanks, cheers!
If you're just calling awk then you don't even need the for loop. Just pass it all of the file names at once.
awk '/whatever/' "$#"
If you do want to loop over all the command-line arguments, write:
for f in "$#"; do
...
done
Or, since in "$#" is implied:
for f; do
...
done
If you want to store them in an intermediate variable, you need to use an array:
victims=("$#")
for f in "${victims[#]}"; do
...
done
Also, you should avoid explicitly invoking bash. Run the script directly so it can use whatever shell's listed in its shebang line.
bash myscript.sh *log*
./myscript.sh *log*
You need to watch out how you call your script. Suppose your script myscript.sh is simply
victims="$1"
echo "$victims"
and your cwd contains files a.log, another.log and logmore.txt.
Then, executing
myscript.sh *log*
Wil result in simply
a.log
because "*log*" is interpreted by the shell before calling myscript.sh. In fact, you're executing
myscript.sh a.log another.log logmore.txt
and your script only handles the first parameter. Also very funny is, when your cwd contains no file with "log" in its name, your script will result in:
*log*
So, your call should be:
myscript.sh "*log*"
and your script should handle the fact that its input may be a regulare expression iso. an existing filename.

Read command line arguments with input redirection operator in bash

I need to read command line arguments. First arg is script name. second one is redirection operator i.e. "<" and third one is input filename. When I tried to use "$#", I got 0. When I used "$*", it gave me nothing. I have to use "<" this operator. My input file consists of all user input data. If I don't use the operator, It asks user for the input. Can someone please help me? Thank you !
Command Line :
./script_name < input_file
Script:
echo "$*" # gave nothing
echo "$#" # gave me 0
I need to read input filename and store it to some variable. Then I have to change the extension of it. Any help/suggestions should be appreciated.
When a user runs:
./script_name <input_file
...that's exactly equivalent to if they did the following:
(exec <input_file; exec ./script_name)
...first redirecting stdin from input_file, then invoking the script named ./script_name without any arguments.
There are operating-system-specific interfaces you can use to get the filename associated with a handle (when it has one), but to use one of these would make your script only able to run on an operating system providing that interface; it's not worth it.
# very, very linux-specific, won't work for "cat foo | ./yourscript", generally evil
if filename=$(readlink /proc/self/fd/0) && [[ -e $filename ]]; then
set -- "$#" "$filename" # append filename to the end of the argument list
fi
If you want to avoid prompting for input when an argument is given, and to have the filename of that argument, then don't take it on stdin but as an argument, and do the redirection yourself within the script:
#!/bin/bash
if [[ $1 ]]; then
exec <"$1" # this redirects your stdin to come from the file
fi
# ...put other logic here...
...and have users invoke your script as:
./script_name input_file
Just as ./yourscript <filename runs yourscript with the contents of filename on its standard input, a script invoked with ./yourscript filename which invokes exec <"$1" will have the contents of filename on its stdin after executing that command.
< is used for input redirection. And whatever is at the right side of < is NOT a command line argument.
So, when you do ./script_name < input_file , there will be zero (0) command line arguments passed to the script, hence $# will be zero.
For your puprpose you need to call your script as:
./script_name input_file
And in your script you can change the extension with something like:
mv -- "$1" "${1}_new_extension"
Edit: This was not what OP wanted to do.
Altough, there is already another spot on answer, I will write this for the sake of completeness. If you have to use the '<' redirection you can do something like this in your script.
while read filename; do
mv -- "$filename" "${filename}_bak"
done
And call the script as, ./script < input_file. However, note that you will not be able to take inputs from stdin in this case.
Unfortunately, if you're hoping to take redirection operators as arguments to your script, you're not going to be able to do that without surrounding your command line arguments in quotes:
./script_name "<input_file"
The reason for this is that the shell (at least bash or zsh) processes the command before ever invoking your script. When the shell interprets your command, it reads:
[shell command (./script_name)][shell input redirection (<input_file)]
invoking your script with quotes effectively results in:
[shell command (./script_name)][script argument ("<input_file")]
Sorry this is a few years late; hopefully someone will find this useful.

beginner's bash script with parameters to be fed to the commands in the script

I want to make a bash script which invokes some simple commands in the following way
./myscript magicword
would invoke
#!/bin/bash
cat * | grep "$#"
However I want to add one more level of complexity, for example triggering case sensitivity or not. Thus I want to include he parsing of arguments such as
./myscript --case-insensitive magicword
which should be interpreted by the script as grep -i "$#".
I have seen many tutorials about bash scripting with arguments, and case structures, but those had to do mostly with each executing different commands altogether, or assigning variables (which are not actually using them as command options).
I didn't find a way of a syntax such as grep $1 "$#" or something, where $1 ought to be null, or anything else.

Can you wrapper each command in GNU's make?

I want to inject a transparent wrappering command on each shell command in a make file. Something like the time shell command. ( However, not the time command. This is a completely different command.)
Is there a way to specify some sort of wrapper or decorator for each shell command that gmake will issue?
Kind of. You can tell make to use a different shell.
SHELL = myshell
where myshell is a wrapper like
#!/bin/sh
time /bin/sh "$0" "$#"
However, the usual way to do that is to prefix a variable to all command calls. While I can't see any show-stopper for the SHELL approach, the prefix approach has the advantage that it's more flexible (you can specify different prefixes for different commands, and override prefix values on the command line), and could be visibly faster.
# Set Q=# to not display command names
TIME = time
foo:
$(Q)$(TIME) foo_compiler
And here's a complete, working example of a shell wrapper:
#!/bin/bash
RESULTZ=/home/rbroger1/repos/knl/results
if [ "$1" == "-c" ] ; then
shift
fi
strace -f -o `mktemp $RESULTZ/result_XXXXXXX` -e trace=open,stat64,execve,exit_group,chdir /bin/sh -c "$#" | awk '{if (match("Process PID=\d+ runs in (64|32) bit",$0) == 0) {print $0}}'
# EOF
I don't think there is a way to do what you want within GNUMake itself.
I have done things like modify the PATH env variable in the Makefile so a directory with my script linked to all name the bins I wanted wrapped was executed rather than the actual bin. The script would then look at how it was called and exec the actual bin with the wrapped command.
ie. exec time "$0" "$#"
These days I usually just update the targets in the Makefile itself. Keeping all your modifications to one file is usually better IMO than managing a directory of links.
Update
I defer to Gilles answer. It's a better answer than mine.
The program that GNU make(1) uses to run commands is specified by the SHELL make variable. It will run each command as
$SHELL -c <command>
You cannot get make to not put the -c in, since that is required for most shells. -c is passed as the first argument ($1) and <command> is passed as a single argument string as the second argument ($2).
You can write your own shell wrapper that prepends the command that you want, taking into account the -c:
#!/bin/sh
eval time "$2"
That will cause time to be run in front of each command. You need eval since $2 will often not be a single command and can contain all sorts of shell metacharacters that need to be expanded or processed.

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