kernel programming serial uart problem

I assume that you do register the interrupts to the kernel correctly.

There are four common address bases for serial chips in PC:

- 0x2e8 - 0x2f8 - 0x3e8 - 0x3f8

The commonly used interrupts are 3 and 4.

For PCI-based interfaces you have to read the addresses from the PCi configuration space. To have a look:

cat /proc/pci

HTH

i do all the necessery work. the fact is that it works in my pc so i register interrupts. i use the 03f8/4 io_base/irq combination. I have tried with serial support on and off from kernel and it works so my driver doesn't infect the kernel if it has serial support.

static int driver_init_module(void){ ... request_region(eib_io, BYTE , eib_dev); //eib_io=0x03f8 BYTE=8 eib_dev=eib0 //eib_irq=4 retval = request_irq(eib_irq,interrupt_handler,SA_INTERRUPT,eib_dev,NULL); if(retval) { release_region(eib_io, BYTE); return retval; } //eib_major=61 experimental eib_node=/dev/eib0 retval = register_chrdev( eib_major , eib_node , &eib_fops ); if(retval

Good ...

I'd next try the 0x2f8 / 3 combination.

You didn't forget to set OUT2 in the MCR register? (that is used to gate the IRQ line on standard PC platforms)

Have you turned on the OUT2 line on the UART? In PC architecture this is used as a gate for the UART interrupt sources. It's bit 3 of the Modem Control Register.

There are a few chipsets that don't honour this scheme - but I think most do.

Regards.

my friend u save me!!!!!!!!!!!!!!!!!!

i spend 1,5 month in that stupid think!!! another time i should read the specs with more attention by the way do u know in what i/o chips the out2 works and in what it don't?