After issuing: function myfunction { echo $1 $2 ; } in Bash, which output does: myfunction A B C Produce?

In Bash scripting, `$1` refers to the first argument passed to a function or script, and `$2` refers to the second argument. The function `myfunction` is defined to `echo` only `$1` and `$2`, so when `myfunction A B C` is executed, it substitutes `A` for `$1` and `B` for `$2`, producing the output "A B". The argument `C` is not captured by any positional parameter in the `echo` command within the function definition.

This would be the output if the function was defined as `echo $@` or `echo $*`, which echoes all arguments.

This output implies the function somehow skipped the second argument and directly accessed the third, which is not how positional parameters work.

This output implies the function accessed the second and third arguments, ignoring the first, which is incorrect for `$1` and `$2`.

This output implies a reversal and access to all arguments, which is inconsistent with `echo $1 $2`.