Hi, I have read a lot of, I think, from the forum but...here is my problem: (TS7800)
I'm working with this code
#include<unistd.h>
#include<sys/types.h>
#include<sys/mman.h>
#include<stdio.h>
#include<fcntl.h>
#include<string.h>
#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
Thanks so much from newbee-world.
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