Multiple Concurrent Coprocesses in Bash – Is It Possible?

If you can guarantee that some character will never occur in the first file then you can use paste.

For example you know for sure that @ will never occur:

paste -d@ file1 file2 | while IFS="@" read -r f1 f2
do
  printf 'f1: %s\n' "$f1"
  printf 'f2: %s\n' "$f2"
done

Note that it is enough if the character is guaranteed to not occur in the first file. This is because read will ignore IFS when filling the last variable. So even if @ occurs in the second file it will not be split.

Example using some bash features for arguably cleaner code and paste using default delimiter tab:

while IFS=$'\t' read -r f1 f2
do
  printf 'f1: %s\n' "$f1"
  printf 'f2: %s\n' "$f2"
done < <(paste file1 file2)

Bash features used: ansi c string ($'\t') and process substitution (<(...)) to avoid the while loop in a subshell problem.

If you cannot be certain that any character will never occur in both files then you can use two file descriptors.

while true
do
  read -r f1 <&3 || break
  read -r f2 <&4 || break
  printf 'f1: %s\n' "$f1"
  printf 'f2: %s\n' "$f2"
done 3<file1 4<file2

Not tested much. Might break on empty lines.

File descriptors number 0, 1, and 2 are already used for stdin, stdout, and stderr, respectively. File descriptors from 3 and up are (usually) free. The bash manual warns from using file descriptors greater than 9, because they are "used internally".

Note that open file descriptors are inherited to shell functions and external programs. Functions and programs inheriting an open file descriptor can read from (and write to) the file descriptor. You should take care to close all file descriptors which are not required before calling a function or external program.

Here is the same program as above with the actual work (the printing) separated from the meta-work (reading line by line from two files in parallel).

work() {
  printf 'f1: %s\n' "$1"
  printf 'f2: %s\n' "$2"
}

while true
do
  read -r f1 <&3 || break
  read -r f2 <&4 || break
  work "$f1" "$f2"
done 3<file1 4<file2

Now we pretend that we have no control over the work code and that code, for whatever reason, tries to read from file descriptor 3.

unknowncode() {
  printf 'f1: %s\n' "$1"
  printf 'f2: %s\n' "$2"
  read -r yoink <&3 && printf 'yoink: %s\n' "$yoink"
}

while true
do
  read -r f1 <&3 || break
  read -r f2 <&4 || break
  unknowncode "$f1" "$f2"
done 3<file1 4<file2

Here is an example output. Note that the second line from the first file is "stolen" from the loop.

f1: file1 line1
f2: file2 line1
yoink: file1 line2
f1: file1 line3
f2: file2 line2

Here is how you should close the file descriptors before calling external code (or any code for that matter).

while true
do
  read -r f1 <&3 || break
  read -r f2 <&4 || break
  # this will close fd3 and fd4 before executing anycode
  anycode "$f1" "$f2" 3<&- 4<&-
  # note that fd3 and fd4 are still open in the loop
done 3<file1 4<file2

Your question is a bit rambling. I'll answer what seems to be the central part, on the difference between ksh and bash that you observe.

You have encountered what is probably the #1 incompatibility between ksh and bash when it comes to scripts. ATT ksh (both ksh88 and ksh93) and zsh execute the last (rightmost) command in a pipeline in the parent shell, whereas other shells (Bourne, ash, bash, pdksh, mksh) execute all the commands including the last one in a subshell.

Here is a simple test program:

msg="a subshell"
true | msg="the parent shell"
echo "This shell runs the last command of a pipeline in $msg"

In ATT ksh and zsh, the second assignment to msg is executed in the parent shell so the effect is visible after the pipeline. In other shells, this assignment is executed in a subshell so the first assignment remains in place in the parent.

A workaround is to execute the rest of the script in the pipeline. This is a common idiom for reading data and doing some processing afterward:

output_some_stuff | {
  var=
  while IFS= read -r line; do
    var=$(process "$line")
  done
  use "$var"
}

You appear to have run into a ksh bug. I recommend upgrading to a non-buggy version. If that isn't possible, try Stephane Chazelas's workaround. While you can try running your scripts in bash, it is not (and does not pretend to be) a drop-in replacement for ksh; there are plenty of ksh features that bash doesn't have (and vice versa). Bash and ksh are only compatible in their POSIX core and some other central features (in particular arrays, [[ … ]], and local variables in functions declared by typeset).

You could also try zsh, which when invoked as ksh behaves in a way that's a bit closer to ksh than bash is. You may nonetheless run into incompatibilities.