What's the difference between <<EOF and <<\EOF heredocs in shell

Question

I have noticed several distinction between them:

Inside a <<EOF heredoc, new values can not be assigned to variables:

bash <<EOF
s=fds
echo $s
EOF

will print empty line, where

bash <<\EOF
s=fds
echo $s
EOF

will print the value of the variable s.

Global variables can be accessed within <<EOF but not within <<\EOF (with export it is possible to access variables inside <<\EOF):

s=fds
bash <<EOF
echo $s
EOF

will print the value fds, where,

s=fds
bash <<\EOF
echo $s
EOF

will print empty line.

So what are the differences between them and what is the legitimate documented behavior?

Answer 1

From the POSIX spec:

If any character in word is quoted, the delimiter shall be formed by performing quote removal on word, and the here-document lines shall not be expanded. Otherwise, the delimiter shall be the word itself.

So the <<EOF version has the shell expand all variables before running the here doc contents and the <<\EOF (or <<'EOF' or <<EO'F' etc.) versions don't expand the contents (which lets bash in this case do that work).

Try it with cat instead of bash for a clearer view on what is happening.

Also with printf '[%s]\n' "$s" and/or possibly bash -x instead of bash:

$ bash -x <<EOF
s=fds
printf '[%s]\n' "$s"
EOF
+ s=fds
+ printf '[%s]\n' ''
[]

$ bash -x <<\EOF
s=fds
printf '[%s]\n' "$s"
EOF
+ s=fds
+ printf '[%s]\n' fds
[fds]

Answer 2

Documentation: http://www.gnu.org/software/bash/manual/bash.html#Here-Documents

In your first example the delimiter is unquoted, so variable expansion occurs and it's like you're running the code

echo "s=fds
echo $s" | bash

which expands $s in the current shell, where it's empty. So the new shell sees

s=fds
echo