How to get the content of a function in a string using bash?

Question

I have already searched about this particular problem, but couldn't find anything helpful.

Let's assume I have following functions defined in my ~/.bashrc (Note: this is pseudo-code!):

ANDROID_PLATFORM_ROOT="/home/simao/xos/src/"

function getPlatformPath() {
  echo "$ANDROID_PLATFORM_ROOT"
}

function addCaf() {
  # Me doing stuff
  echo "blah/$(getPlatformPath)"
}

function addAosp() {
  # Me doing stuff
  echo "aosp/$(getPlatformPath)"
}

function addXos() {
  # Me doing stuff
  echo "xos/$(getPlatformPath)"
}

function addAllAll() {
  cd $(gettop)
  # repo forall -c "addCaf; addAosp; addXos" # Does not work!
  repo forall -c # Here is where I need all those commands
}

My problem:

I need to get the functions addCaf, addAosp and addXos in one single line.

Like you can run following in bash (pseudo code):

dothis; dothat; doanotherthing; trythis && succeedsdothis || nosuccessdothis; blah

I would like to run all commands inside the three functions addCaf, addAosp and addXos in just one line.

Any help is appreciated.

What I already tried:

repo forall -c "bash -c \"source ~/.bashrc; addAllAll\""

But that didn't work as well.

Edit:

To clarify what I mean.

I want something like that as a result:

repo forall -c 'function getPlatformPath() { echo "$ANDROID_PLATFORM_ROOT"; }; ANDROID_PLATFORM_ROOT="/home/simao/xos/src/"; echo "blah/$(getPlatformPath)"; echo "aosp/$(getPlatformPath)"; echo "xos/$(getPlatformPath)"'

But I don't want to write that manually. Instead, I want to get those lines from the functions that already exist.

Not sure what you are asking, but if you want to create a function on one line, you can use function f() { echo hi; }. Note the final ;. — Mark Setchell
– Mark Setchell, Commented Aug 3, 2016 at 11:46
Well, I want to get the content of a function in one line in one string. — xdevs23
– xdevs23, Commented Aug 3, 2016 at 12:11
Why do you need the content of the functions instead of just running the functions themselves? Because you need to embed the calls in the argument to repo (whatever that is)? Does type addXos do (most of) what you need? Or actually declare -f addXos? — Etan Reisner
– Etan Reisner, Commented Aug 3, 2016 at 12:23
Because, for whatever reason, the repo tool does not accept that. I could be writing the stuff into the quotes but I am lazy :P — xdevs23
– xdevs23, Commented Aug 3, 2016 at 12:37
From your question, it's unclear what repo all -c does with the next positional argument. Does it run shell substitution on it (like bash -c), or does it interpret it as a command? The last 2 code snippets suggest different things. — Matei David
– Matei David, Commented Aug 3, 2016 at 14:35

Vicente Olivert Riera · Accepted Answer · 2016-08-03 12:52:46Z

6

You can use type and then parse its output to do whatever you want to do with the code lines.

$ foo() {
> echo foo
> }

$ type foo
foo is a function
foo () 
{ 
    echo foo
}

Perhaps this example makes things more clear:

#!/bin/bash

foo() {
    echo "foo"
}

bar() {
    echo "bar"
}

export IFS=$'\n'

for f in foo bar; do
    for i in $(type $f | head -n-1 | tail -n+4); do
        eval $i
    done
done

exit 0

This is how it looks:

$ ./funcs.sh 
foo
bar

What the script is doing is first loop over all the functions you have (in this case only foo and bar). For each function, it loops over the code of that function (skipping the useless lines from type's output) and it executes them. So at the end it's the same as having this code...

echo "foo"
echo "bar"

...which are exactly the code lines inside the functions, and you are executing them one after the other.

Note that you could also build a string variable containing all the code lines separated by ; if instead of running eval on every line you do something like this:

code_lines=

for f in foo bar; do
        for i in $(type $f | head -n-1 | tail -n+4); do
                if [ -z $code_lines ]; then
                        code_lines="$i"
                else
                        code_lines="${code_lines}; $i"
                fi  
        done
done

eval $code_lines

edited Aug 3, 2016 at 12:52

answered Aug 3, 2016 at 12:26

Vicente Olivert Riera

1,27010 silver badges15 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

xdevs23 Over a year ago

That's almost what I need. Can you also tell me how I can remove foo is a function and foo() { }, only keeping the lines themselves in one single line like this: echo foo; echo foo2; echo foo3;..., assuming that those commands are in that particular function?

Matei David · Accepted Answer · 2016-08-04 14:31:17Z

2

Assuming that repo forall -c interprets the next positional argument just as bash -c, try:

foo () { 
    echo "foo!"
}
boo () { 
    if true; then
        echo "boo!"
    fi
}
echo works | bash -c "source "<(typeset -f foo boo)"; foo; boo; cat"

Note:

The difference from the original version is that this no longer interferes with stdin.
The <(...) substitution is unescaped because it must be performed by the original shell, the one where foo and boo are first defined. Its output will be a string of the form /dev/fd/63, which is a file descriptor that is passed open to the second shell, and which contains the forwarded definitions.

edited Aug 4, 2016 at 14:31

answered Aug 3, 2016 at 14:40

Matei David

2,3823 gold badges23 silver badges37 bronze badges

Comments

agc · Accepted Answer · 2016-08-04 19:06:13Z

1

Make a dummy function foo(), which just prints "bar":

foo() { echo bar ; }

Now a bash function to print what's in one (or more) functions. Since the contents of a function are indented with 4 spaces, sed removes any lines without 4 leading spaces, then removes the leading spaces as well, and adds a ';' at the end of each function:

# Usage: in_func <function_name1> [ ...<function_name2> ... ]
in_func() 
    { while [ "$1" ] ; do \
          type $1 | sed  -n '/^    /{s/^    //p}' | sed '$s/.*/&;/' ; shift ; \
      done ; }

Print what's in foo():

in_func foo

Output:

echo bar;

Assign what's in foo() to the string $baz, then print $baz:

baz="`in_func foo`" ; echo $baz

Output:

echo bar;

Run what's in foo():

eval "$baz"

Output:

bar

Assign what's in foo() to $baz three times, and run it:

baz="`in_func foo foo foo`" ; eval "$baz"

Output:

bar
bar
bar

edited Aug 4, 2016 at 19:06

answered Aug 3, 2016 at 15:41

agc

8,5342 gold badges33 silver badges53 bronze badges

5 Comments

agc Over a year ago

Caveat: some folks regard most any language's form of eval as eeville, not entirely without reason.

xdevs23 Over a year ago

This works. Just copy&pasted that in_func function, then just ran the following: repo forall -c "$(in_func addAosp) $(in_func addXos) $(in_func addCaf). But it is a bit buggy with the repo tool. Well, this answer is the correct one.

Matei David Over a year ago

Removing spacing and adding ;-s won't work with higher bash constructs such as while, for, if. E.g.: boo () { while false; do echo bar; done; echo ok; }; eval $(in_func boo) gives a syntax error. (The problem is that in_func adds a ; right after do.) My answer somewhere below is not limited in this way, but alas this was accepted.

agc Over a year ago

@MateiDavid, thanks for the bug report. I should have tried in_func in_func, since it too contains a while. Have now added a do check to the answer sed code.

agc Over a year ago

Revised do correction -- just stick a ; on the end only. Should fix the problems MateiDavid pointed out. (Couldn't figure out how to do it without invoking sed twice.)

John Kugelman · Accepted Answer · 2016-08-03 12:26:54Z

1

Shell functions aren't visible to child processes unless they're exported. Perhaps that is the missing ingredient.

export -f addCaf addAosp addXos
repo forall -c "addCaf; addAosp; addXos"

edited Aug 3, 2016 at 12:26

answered Aug 3, 2016 at 12:20

John Kugelman

365k70 gold badges555 silver badges600 bronze badges

1 Comment

xdevs23 Over a year ago

Even though your answer might be correct, I am facing following error message: Got an error, terminating the pool: OSError: [Errno 2] No such file or directory. Seems to be related to the repo tool. Anyways, this does not help me in this particular situation. Do you know how I can echo the content of a function, means all single commands?

Jawad · Accepted Answer · 2016-08-03 11:37:16Z

0

This should work:

repo forall -c "$(addCaf) $(addAosp) $(addXos)"

answered Aug 3, 2016 at 11:37

Jawad

4,7051 gold badge30 silver badges29 bronze badges

2 Comments

xdevs23 Over a year ago

I don't want the result of the functions, I want the content.

xdevs23 Over a year ago

No problem, nice suggestion though.

Collectives™ on Stack Overflow

How to get the content of a function in a string using bash?

5 Answers 5

1 Comment

Comments

5 Comments

1 Comment

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

Comments

5 Comments

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related