'find -exec' a shell function in Linux

Question

Since only the shell knows how to run shell functions, you have to run a shell to run a function. You also need to mark your function for export with export -f, otherwise the subshell won't inherit them:

export -f dosomething
find . -exec bash -c 'dosomething "$0"' {} \;

find . | while read file; do dosomething "$file"; done

Jac's answer is great, but it has a couple of pitfalls that are easily overcome:

find . -print0 | while IFS= read -r -d '' file; do dosomething "$file"; done

This uses null as a delimiter instead of a linefeed, so filenames with line feeds will work. It also uses the -r flag which disables backslash escaping, and without it backslashes in filenames won't work. It also clears IFS so that potential trailing white spaces in names are not discarded.

Add quotes in {} as shown below:

export -f dosomething
find . -exec bash -c 'dosomething "{}"' \;

This corrects any error due to special characters returned by find, for example files with parentheses in their name.

Processing results in bulk

For increased efficiency, many people use xargs to process results in bulk, but it is very dangerous. Because of that there was an alternate method introduced into find that executes results in bulk.

Note though that this method might come with some caveats like for example a requirement in POSIX-find to have {} at the end of the command.

export -f dosomething
find . -exec bash -c 'for f; do dosomething "$f"; done' _ {} +

find will pass many results as arguments to a single call of bash and the for-loop iterates through those arguments, executing the function dosomething on each one of those.

The above solution starts arguments at $1, which is why there is a _ (which represents $0).

Processing results one by one

In the same way, I think that the accepted top answer should be corrected to be

export -f dosomething
find . -exec bash -c 'dosomething "$1"' _ {} \;

This is not only more sane, because arguments should always start at $1, but also using $0 could lead to unexpected behavior if the filename returned by find has special meaning to the shell.

Have the script call itself, passing each item found as an argument:

#!/bin/bash

if [ ! $1 == "" ] ; then
   echo "doing something with $1"
   exit 0
fi

find . -exec $0 {} \;

exit 0

When you run the script by itself, it finds what you are looking for and calls itself passing each find result as the argument. When the script is run with an argument, it executes the commands on the argument and then exits.

For those of you looking for a Bash function that will execute a given command on all files in current directory, I have compiled one from the above answers:

toall(){
    find . -type f | while read file; do "$1" "$file"; done
}

Note that it breaks with file names containing spaces (see below).

As an example, take this function:

world(){
    sed -i 's_hello_world_g' "$1"
}

Say I wanted to change all instances of "hello" to "world" in all files in the current directory. I would do:

toall world

To be safe with any symbols in filenames, use:

toall(){
    find . -type f -print0 | while IFS= read -r -d '' file; do "$1" "$file"; done
}

(but you need a find that handles -print0 e.g., GNU find).