define a shell function without a shell script in bash?

Question

All I want to do is write and export a shell function without:

1) opening a file
2) editing it
3) saving it
4) running it

I'm aware of the implications... Something like:

$ afunc () { echo "i am awesome" } && export -f afunc

When i call afunc it will print "i am awesome" but if i try to do that I get this situation

$ afunc () { echo "aaa" }
>

Anyway way i can do this dynamically from stdin or something?

sorry about dupe, did not see those answers – niken Aug 31 '17 at 19:23 — niken, Aug 31 '17 at 19:23

score 2 · Accepted Answer · answered Aug 31 '17 at 18:14

This isn't a problem with being inside/outside a script, but a problem in how you're compressing your definition down to a one-liner: You need (but are not including) a semicolon before the closing brace.

The following works:

afunc () { echo "i am awesome"; } && export -f afunc

define a shell function without a shell script in bash?

1 Answers1