[ragel-users] Re: Bug or feature?

[Adrian Thurston]

Carlos Antunes wrote:
> machine_a = machine_b %action_b machine_c
> 
> Now, this is semantically equivalent to (assuming that ">" is really
> associated with the start state:
> 
> machine_a = machine_b  machine_c >action_b

They are not quite equivalent. Consider the case of machine_c accepting
the zero-length word (start state is final). Then action_b will be a
pending out action of machine_a.

Note however that the first form above is equivalent to the second form
under the old semantics of >.

One thing I don't like about the syntax you propose is that it creates
an ambiguity in the ragel language itself. When you parse % is it
associated with the tree on the left or the tree on the right? Right now
all embedding operators are currently of this form:

<expr> <op> <action>

and this syntax changes that.

Also note that you can achieve the same thing using the existing
language by putting "" in front of %. Like this:

machine_a = machine_b ( ""%action_b machine_c );

So that brings up an interesting point. The line directly above is
another way to emulate the old semantics of >.

-Adrian

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