Redirect expect stdin to spawn call - bash

Question

Is there a way to redirect stdin sent to expect such that it is fed to a spawn call within expect? In my example below I am embedding expect within a shell function to which I want to pipe another shell script via heredoc and supressing the output by capturing it to a shell variable.
psshstdin() {
local user=$1 pass=$2 hosts=$3
out=$(expect -c '
set timeout 15
spawn pssh -i -h '"$hosts"' -p 100 -l '"$user"' -A -o ./ -x-oStrictHostKeyChecking=no <EXPECT_STDIN_HERE
expect "assword:" { send '\""$pass\r\""' }
interact
'<<EOF
echo "hello"
echo "world"
EOF
)
}
SOLUTION: I had to post this here since I don't have enough reputation points to answer my own question so quickly.
I was able to resolve it by trying the same techniques applied in this issue. I didn't think that solution was applicable initially, but it was. The working code is shown below.
psshstdin() {
local user=$1 pass=$2 hosts=$3
out=$(expect -c '
set timeout 30
spawn pssh -I -h '"$hosts"' -p 100 -l '"$user"' -A -o ./ -x-oStrictHostKeyChecking=no
while {[gets stdin line] != -1} {
send "$line\n"
}
send \004
expect "assword:" { send '\""$pass\r\""' }
expect {
"END_TOKEN_OF_SCRIPT" {
exit 0
}
default {
exit 1
}
}'<&0)
}
I can call it with something like:
psshstdin myusername mypassword ssh_hosts_file<<EOF
echo "hello"
echo "world"
EOF

Answer 1

You can capture stdin to a variable with stdin=$(cat -)
Update: let expect collect the stdin:
untested, but perhaps:
psshstdin() {
local user=$1 pass=$2 hosts=$3
out=$(expect -c '
set stdin [read stdin]
puts "debug: sending the following stdin to pssh:\n$stdin\n--END--"
set timeout 15
spawn echo "$stdin" | pssh -i -h '"$hosts"' -p 100 -l '"$user"' -A -o ./ -x-oStrictHostKeyChecking=no
expect "assword:" { send "'"$pass"'\r" }
interact
'<<EOF
echo "hello"
echo "world"
EOF
)
}