Bash/sh - difference between && and ;

Question

If previous command failed with ; the second one will run.

But with && the second one will not run.

This is a "lazy" logical "AND" operand between operations.

I'm using && because a long time ago at the nearby computer:

root# pwd
/
root# cd /tnp/test; rm -rf *
cd: /tnp/test: No such file or directory
...
... and after a while ...
...   
^C

but not helped... ;)

cd /tnp/test && rm -rf * is safe... ;)

In cmd1 && cmd2, cmd2 is only executed if cmd1 succeeds (returns 0).

In cmd1 ; cmd2, cmd2 is executed in any case.

Both constructs are part of a POSIX-compliant shell.

&& means to execute next command if the previous exited with status 0. For the opposite, use || i.e. to be executed if previous command exits with a status not equal to 0 ; executes always.

Very useful when you need to take a particular action depending on if the previous command finished OK or not.

Commands separate by ; are executed sequentially regardless of their completion status.

With &&, the second command is executed only if the first completes successfully (returns exit status of 0).

This is covered in the bash manpage under Lists. I would expect any Unix-like shell to support both of these operators, but I don't know specifically about the Android shell.

&& allows for conditional execution while ; always has the second command being executed.

In e.g. command1 && command2, command2 will only execute when command1 has terminated with exit 0, signalling all went well, while in command1 ; command2 the second command will always be executed no matter what the result was of command1.

&& is logical AND in bash. Bash has short-circuit evaluation of logical AND. This idiom is a simpler way of expressing the following:

cmd1;rc=$?
if [ $rc -eq 0 ]; then
   cmd2
fi

Whereas the ; version is merely:

cmd1
cmd2