java,enums,finite-state-machine,type-2-dimension

I have a little trouble understanding what you are trying to do, but might this do it? enum States { IDLE, NEAREST, NEARESTTOHQ,NORTH,SOUTH,EAST,WEST,IDLESPAWN } public enum Modules { ATTACK(IDLE, NEAREST, NEARESTTOHQ), MOVE(IDLE, NORTH, SOUTH, EAST, WEST), SPAWN(IDLE, SIMPLESPAWN); private States[] states; public Modules(States... states) { this.states=states; } } (Note that...

java,return,finite-state-machine

In this case it's mainly a formatting alternative to a series of else if clauses. It is logically equivalent to if (<condition>) { <code> } else if (<condition>) { <code> } else { <code> } ...

persistence,workflow,finite-state-machine,state-machine-workflow

It looks like what you need to do is construct your StateMachine with a custom accessor and mutator, something like this: public class PersistentMutator<S> implements Action1<S> { Foo foo = null; @Inject FooRepository fooRepository; public PersistentMutator(Foo foo) { this.foo = foo; } @Override public void doIt(S s) { foo.setState(s); fooRepository.save(foo)...

haskell,finite-state-machine,transducer

A Mealy machine alternately reads an a from a stream of inputs a and outputs a b to a stream of outputs. It reads first and then outputs once after each read. newtype Mealy a b = Mealy { runMealy :: a -> (b, Mealy a b) } A Moore...

state-machines,fsm,finite-state-machine

I am sure there is no single right answer here. But I would say it depends. If you want to consider the whole Product a single FSM, then your state would be a sum of all the attributes. Such as sellable, taxable, tangible, ... At one point it might be...

With a sane libc you would use #include <ctype.h> ... if (!isprint((int)ch) { unsigned x = ch; printf ("[0x%02x]", 0xff&(int)ch); } ... to find non-printable ascii characters, assumed that char ch is your current input character. To use them in a command line you could use printf(1) from the command...

To show what I meant by passing a delegate: class MyStateMachine { private readonly Func<string> askForName; public MyStateMachine(Func<string> askForName) { this.askForName = askForName; } // ... void StateTransitionForActionX() { var name = askForName(); // ... } } public MyStateMachine CreateMachine() { return new MyStateMachine ( () => { Console.WriteLine("Please, enter...

verilog,system-verilog,fsm,finite-state-machine,precedence

Ideally, you'd need to cover all the possible combinations of input events in each state to get a proper DFA (deterministic FSM). However, you can get away by fully specifying the triggers in terms of input signals, and let your HDL default to "no transition". In that case: transition from...

if-statement,vhdl,finite-state-machine

No your syntax isn't correct. As noted by Amir : process(clk,adsn,blastn,lwdrn,lhold,adds_in,adds_4msb) begin variable sa:std_logic:='0'; variable a31_a28 :std_logic_vector(3 downto 0):="0000"; variable temp:std_logic_vector(9 downto 2):="00000000"; Should be: process(clk,adsn,blastn,lwdrn,lhold,adds_in,adds_4msb) variable sa:std_logic:='0'; variable a31_a28 :std_logic_vector(3 downto 0):="0000"; variable temp:std_logic_vector(9 downto 2):="00000000"; begin begin separates the process declarative part from the process statement part here....

algorithm,computer-science,finite-state-machine

No. Intuitively, an algorithm can only be represented as an FSM if it uses only a finite amount of state. For instance, you couldn't sort an arbitrary-length list with an FSM. Now, add an unbounded amount of state to an FSM -- like an infinite one-dimensional array of values... and...

c,switch-statement,finite-state-machine

It is a pretty much straight forward question and the solution is also very simple. I have 7 states namely from 0 to 6 as shown by diagram.0 is the initial state. 3,4,5 could be the final states and 6 is the dead state. state 0: Initial state and from...

racket,fsm,dfa,finite-state-machine,nfa

Your plan does not work. Consider this machine: States: S,L,F Alphabet: a,b Start state: S Final state S S-a->L (arrow from S to L given a) S-b->F L-a->L Strings accepted by this machine: {b} If you make a new machine where L is a final state, then {empty,a,aa,aaa,...} will be...

io,complexity-theory,finite-state-machine,automaton

For a Deterministic FSM (i.e., one without epsilon state transitions), there is a unique input sequence leading to the state if and only if the following conditions are met: 1) There must exist a path to the state. (An isolated unreachable state could not qualify). 2) There is no path...