What does the combination ^foo^bar do in Bash?
Bash calls this a quick substitution. It's in the "History Expansion" section of the Bash man page, under the "Event Designators" section (online manual):
^string1^string2^
Quick substitution. Repeat the previous command, replacing string1 with string2. Equivalent to
!!:s/string1/string2/
So ^foo^bar
would run the previously executed command, but replace the first occurence of foo
with bar
.
Note that for s/old/new/
, the bash man page says "The final
delimiter is optional if it is the last character of the
event line." This is why you can use ^foo^bar
and aren't required to use ^foo^bar^
.
(See this answer for a bunch of other designators, although I didn't mention this one there).
^foo^bar
executes that last command, replacing the first instance of foo
with bar
. For example:
$ ech "hello"
-bash: ech: command not found
$ ^ech^echo
echo "hello"
hello