Using Return in bash

I'm wanting this script to print 1,2,3.. without the use of functions, just execute two.sh then carry on where it left off, is it possible?

[root@server:~]# cat testing.sh                                  
#!/bin/bash
echo "1"
exec ./two.sh
echo "3"

[root@server:~]# cat two.sh                                     
#!/bin/bash
echo "2"
return

exec, if you give it a program name ^a, will replace the current program with whatever you specify.

If you want to just run the script (in another process) and return, simply use:

./two.sh

to do that.

For this simple case, you can also execute the script in the context of the current process with:

. ./two.sh

That will not start up a new process but will have the side-effect of allowing two.sh to affect the current shell's environment. While that's not a problem for your current two.sh (since all it does is echo a line), it may be problematic for more complicated scripts (for example, those that set environment variables).

^a Without a program name, it changes certain properties of the current program, such as:

exec >/dev/null

which simply starts sending all standard output to the bit bucket.

Sure, just run:

echo "1"
./two.sh
echo "3"