bash recursive xtrace

Question

Is there any way to run bash script X so that if X call executable bash script Y then Y starts by 'sh -eux'?

X.sh:

./Y.sh

Y.sh:

#!/bin/sh
echo OK

Answer 1

So i've got tool to debug shell scripts:

  #!/bin/sh

  # this tool allows to get full xtrace of any shell script execution
  # only argument is script name


  out=out.$1
  err=err.$1
  tmp=tmp.$1

  echo "export SHELLOPTS; sh $@ 1>> $out 2>> $err" > $tmp
  sh -eux $tmp &> /dev/null
  rm $tmp

Answer 2

It is possible to make a subshell run using the same shell options set in the parent by exporting the SHELLOPTS environment variable.

In your case where X.sh and Y.sh cannot be edited, I'd create a wrapper script that simply exports SHELLOPTS before calling X.sh.

Example:

#!/bin/sh
# example X.sh which calls Y.sh
./Y.sh

.

#!/bin/sh
# example Y.sh which needs to be called using sh -eux
echo $SHELLOPTS

.

#!/bin/sh -eux
# wrapper.sh which sets the options for all sub shells
export SHELLOPTS
./X.sh

Calling X.sh directly shows that -eux options are not set in Y.sh

[lsc@aphek]$ ./X.sh 
braceexpand:hashall:interactive-comments:posix

Calling it via wrapper.sh shows that the options have propagated to the subshells.

[lsc@aphek]$ ./wrapper.sh 
+ export SHELLOPTS
+ ./x.sh
+ ./y.sh
+ echo braceexpand:errexit:hashall:interactive-comments:nounset:posix:xtrace
braceexpand:errexit:hashall:interactive-comments:nounset:posix:xtrace

Tested on GNU bash, version 3.00.15(1)-release. YMMV.