With this modification, the code
> // state = *PBDR; not used now
> if (oldstate != *PBDR)
> {
> printf ("State:%o\n", *PBDR);
prints all the values in the registers.
But, the other question, somebody could explain me this line?
*GPIOBDB = 0x01; //enable debounce on bit 0
(Sorry for the octal mode, it was a proof, must be %x)
naturalwatt escribió:
>
> I have pretty much identical code and it works fine for me.
>
> However you say you can only see two hexadecimal chars.
>
> The variable 'state' is a char, it's 8 bits wide, hence two hex chars. Why would you expect 4? And you are printing in Octal, which might confuse you.
>
> I must admit I'm puzzled why the PBDR variables are int *, not char *, but my code is the same.
>
> --- In salva g ...> wrote:
> >
> >
> > Hi, I have read a lot of, I think, from the forum but...here is my problem: (TS7800)
> > I'm working with this code
> >
> > #includeh>
> > #includetypes.h>
> > #includemman.h>
> > #includeh>
> > #includeh>
> > #includeh>
> >
> > #define DIOBASE 0xE8000000
> > int main(int argc, char **argv)
> > {
> > volatile unsigned int *PBDR, *PBDDR, *GPIOBDB;
> >
> > int i;
> > unsigned char state;
> > unsigned char *start;
> > int fd = open("/dev/mem", O_RDWR|O_SYNC);
> >
> > start = mmap(0, getpagesize(), PROT_READ|PROT_WRITE, MAP_SHARED, fd,DIOBASE);
> >
> > PBDR = (unsigned int *)(start + 0x04); //port b
> > //starting address of DIO read, E8000000 + 4 = E80 000 004
> > PBDDR = (unsigned int *)(start + 0x14); //port b direction
> >
> > GPIOBDB = (unsigned int *)(start + 0xC4); // debounce on port b
> >
> > *PBDDR = 0x00; // all(8) inputs
> > *GPIOBDB = 0x01; //enable debounce on bit 0
> > state = *PBDR; // read initial state
> >
> > printf ("Press buttons on DIO inputs. Ctrl-C to Quit.\n");
> >
> > int count = 0;
> > unsigned char oldstate = (unsigned char)0;
> > while (1) //(state & 0x01)
> > {
> > state = *PBDR; // remember bit 0 is pulled up with 4.7k ohm
> > if (oldstate != state)
> > {
> > printf ("State:%o\n", state);
> > oldstate = state;
> > }
> > usleep(1000);
> >
> > }
> > printf ("\nDONE\n");
> > close(fd);
> > return 0;
> > }
> >
> > but only have an answer in the shell when I short with ground (pin 2) the 1,3,5, or 7 pin DIO.
> > shorting the other 9,11,13,15, nothing happens, but with peek32 0xE8000004, I can see the changes, the problem is that the program shows only 2 hexadecimal characters. If I could see 4 of them, it will be solved.
> > And, I don't know why doesn't matter if I put (*PBDDR = 0x00 //all inputs) or (*PBDDR = 0xf0;): the program works equal in both.
> > Also, somebody could explain me the line: *GPIOBDB = 0x01; //enable debounce on bit 0
> >
Entra al Nuevo Canal Motor y descubre por qué los coches más rápidos sólo aparcan en MSN. Nuevo diseño, más completo y abierto a tu opinión. ¡Nuevo Canal Motor!
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