cassemblyx86osdevata

What does the function insl do in Os Dev's PCI IDE tutorial?


This is the function that call's insl.

void ide_read_buffer(unsigned char channel, unsigned char reg, unsigned int buffer,
                     unsigned int quads) {
   /* WARNING: This code contains a serious bug. The inline assembly trashes ES and
    *           ESP for all of the code the compiler generates between the inline
    *           assembly blocks.
    */
   if (reg > 0x07 && reg < 0x0C)
      ide_write(channel, ATA_REG_CONTROL, 0x80 | channels[channel].nIEN);
   asm("pushw %es; movw %ds, %ax; movw %ax, %es");
   if (reg < 0x08)
      insl(channels[channel].base  + reg - 0x00, buffer, quads);
   else if (reg < 0x0C)
      insl(channels[channel].base  + reg - 0x06, buffer, quads);
   else if (reg < 0x0E)
      insl(channels[channel].ctrl  + reg - 0x0A, buffer, quads);
   else if (reg < 0x16)
      insl(channels[channel].bmide + reg - 0x0E, buffer, quads);
   asm("popw %es;");
   if (reg > 0x07 && reg < 0x0C)
      ide_write(channel, ATA_REG_CONTROL, channels[channel].nIEN);
}

The link is here, https://wiki.osdev.org/PCI_IDE_Controller#Commands.


Solution

  • What does the function insl do in Os Dev's PCI IDE tutorial?

    void ide_read_buffer(... unsigned int buffer ...)
    

    It is clear why you don't understand the code:

    unsigned int buffer is obviously a bug in the tutorial code. It should be unsigned int * buffer.

    Now it is clear what insl does: It reads quad times from the port given in the first argument and writes the results to the array buffer.

    The black-box behaviour of the function could be explained the following way:

    void insl(unsigned reg, unsigned int *buffer, int quads)
    {
        int index;
        for(index = 0; index < quads; index++)
        {
            buffer[index] = inl(reg);
        }
    }