Why can't I eval inside a while loop?

Question

I want to create bash aliases within a loop where the loop is reading lines from a command. In order to read output line-by-line, I believe I need to pipe the output into read. When I do that, however, the aliases don't exist.

If I include the following in my .bashrc:

for x in a1 a2; do
    eval "alias $x='echo foo'"
done

echo -e "a3\na4" | while read x; do
    eval "alias $x='echo foo'"
done

Aliases a1 and a2 exist, but a3 and a4 do not. What is the difference between those two loops?

Answer 1

The while loop in the pipe runs in a subshell. You can do:

while read x; do
    eval "alias $x='echo foo'"
done << EOF
a3
a4 
EOF

Answer 2

The problem is the pipeline. In a pipeline of the form a | b | c, each of the individual commands a, b, and c is run in a separate subshell [ref], which means that it receives a copy of the parent's execution environment (including aliases), and any changes it makes to its own copy (such as by running alias) will have no effect on the parent [ref].

In your case, you could fix this by writing:

while read x; do
    eval "alias $x='echo foo'"
done < <(echo -e "a3\na4")

which will still run echo -e "a3\na4" in a subshell, but will run the while-loop in the normal/parent execution environment.

Answer 3

In bash 4.2 and later, you can set the lastpipe option to keep the last command of a pipe from executing in a subshell. Job control must also be off (set +m) for this to work.

shopt -s lastpipe
set +m
echo -e "a3\na4" | while read x; do
    alias $x='echo foo'
done