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I know when double exclamation is printed, it executes the previous command. But echo !! gives some strange results which I don't understand. For example when typed below command in bash script, it prints echo too as part of the output
echo $$ echo !! This prints the below output: echo echo $$ echo 3150 (Why does echo precede every output ?) 
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So if you type: some command echo !! the !! is replaced with the contents of the previous command. So it displays and then executes echo some command. Follow this answer to receive notifications.
Bash maintains the history of the commands executed in the current session. We can use the exclamation mark (!) to execute specific commands from the history.
The origin of != is the C family of programming languages, in which the exclamation point generally means "not". In bash, a ! at the start of a command will invert the exit status of the command, turning nonzero values to zero and zeroes to one.
In addition to using single quotes for exclamations, in most shells you can also use a backslash \ to escape it.
When you use history substitution, the shell first displays the command that it's about to execute with all the substitutions shown, and then executes it. This is so you can see what the resulting command is, to confirm that it's what you expected.
So if you type:
some command echo !! the !! is replaced with the contents of the previous command. So it displays and then executes
echo some command It's caused by history expansion. Quote it instead:
echo '!!' echo \!\! Or disable history expansion:
shopt -u -o histexpand ## Or set +H See History Expansion.
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