Use find -exec in a script

Question

I'm a long-time find and bash user, but this seems to be the first time I've tried to run a find command in a script.

I'm trying to loop through all the project source files to find the first argument passed to a few functions. I've written and tested a little awk script that seems to do the job and I can run it from the command line via

find sources -name '*.cpp' -exec awk -f foo.awk {} \;

The trouble comes when I try to put it in a script, via

echo find sources -name *.cpp -exec awk -f foo.awk {} \\\; >foo.sh

so that foo.sh contains

find sources -name *.cpp -exec awk -f foo.awk {} \;

If I now source it into my bash shell via

source foo.sh

I get the error

find: missing argument to `-exec'

I thought that was a quoting issue, so I doubled the backslash before the final semi-colon. It made no difference. Just for laughs, I tried trebling the backslash. Still no difference.

Full disclosure: I'm running on Windows 10, under cygwin, using GNU Awk 5.1.0. I've been writing the scripts by using cat, so I should be clear of any problems with line endings

Update

Except that I don't. Running with a different awk script, the script that I've just pasted into this question works fine. I shall try again tomorrow and post my findings.

Thanks for your forbearance.

Answer 1

This is almost certainly because you're not quoting the *.cpp so it is being expanded before being passed to find or, if you have *cpp files in your current directory, before even writing to the script.

If I'm right, this should do what you need:

echo "find sources -name '*.cpp' -exec awk -f foo.awk {} \;" >foo.sh