How does one determine the page frame number for device memory?

Question

From LDD3/ Ch. 15/ sections "Using remap_pfn_range" and "A Simple Implementation", pfn has been equated to the vm_pgoff field. I am confused by this. How can that be so?

Note that vm_pgoff is described as:

The offset of the area in the file, in pages. When a file or device is mapped, this is the file position of the first page mapped in this area.

Thus if the first page mapped corresponds to the first page of the file as well (which, I think would be quite common), vm_pgoff would be 0. correct? If so, this doesn't seem to be the correct value for the pfn parameter of remap_pfn_range( ). What am I missing here? What is the correct value? For ease of reference, I am reproducing the relevant code from LDD3 below (Page no. 426)

static int simple_remap_mmap(struct file *filp, struct vm_area_struct *vma)
{
if (remap_pfn_range(vma, vma->vm_start, vm->vm_pgoff,
                    vma->vm_end - vma->vm_start,
                    vma->vm_page_prot))
    return -EAGAIN;
...
}

Answer 1

The specific example you've provided is implementing a character device file that allows one to map physical memory, very similar to /dev/mem. By specifying the offset of the file you specify the physical memory address. Hence the calculation that takes the offset and divide in the page size to find the PFN.

For a "real" device driver, you would normally have the physical address of the device memory mapped registers or RAM hard coded from the device specification, and use that to derive the PFN (by dividing by the page size).