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Created February 1, 2012 14:55
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State machines are very simple in C if you use function pointers.
State machines are very simple in C if you use function pointers.
Basically you need 2 arrays - one for state function pointers and one for state
transition rules. Every state function returns the code, you lookup state
transition table by state and return code to find the next state and then
just execute it.
int entry_state(void);
int foo_state(void);
int bar_state(void);
int exit_state(void);
/* array and enum below must be in sync! */
int (* state)(void)[] = { entry_state, foo_state, bar_state, exit_state};
enum state_codes { entry, foo, bar, end};
enum ret_codes { ok, fail, repeat};
struct transition {
enum state_codes src_state;
enum ret_codes ret_code;
enum state_codes dst_state;
/* transitions from end state aren't needed */
struct transition state_transitions[] = {
{entry, ok, foo},
{entry, fail, end},
{foo, ok, bar},
{foo, fail, end},
{foo, repeat, foo},
{bar, ok, end},
{bar, fail, end},
{bar, repeat, foo}};
#define EXIT_STATE end
#define ENTRY_STATE entry
int main(int argc, char *argv[]) {
enum state_codes cur_state = ENTRY_STATE;
enum ret_codes rc;
int (* state_fun)(void);
for (;;) {
state_fun = state[cur_state];
rc = state_fun();
if (EXIT_STATE == cur_state)
cur_state = lookup_transitions(cur_state, rc);
I don't put lookup_transition() function as it is trivial.
That's the way I do state machines for years.
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nmandery commented Aug 6, 2020

Hey everybody, I just saw all of your questions regarding this gist for the first time - sorry for being late to reply. I did not author this code, please see the link to the stackoverflow post in the first line. The original author can be asked there.

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