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Explained why the magic number used in 24bit ROM frame mapping.

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Signed-off-by: Ricky Zhang <rickyzhang@gmail.com>
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Ricky Zhang 2020-07-04 14:45:21 -04:00
parent e63c6b91dc
commit bae7a4d16b
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BasiliskII/src/uae_cpu/memory.cpp
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@ -473,6 +473,34 @@ static void REGPARAM2 fram24_lput(uaecptr, uae_u32) REGPARAM;
static void REGPARAM2 fram24_wput(uaecptr, uae_u32) REGPARAM;
static void REGPARAM2 fram24_bput(uaecptr, uae_u32) REGPARAM;
/*
* Q: Why the magic number 0xa700 and 0xfc80?
*
* A: The M68K CPU used by the earlier Macintosh models such as
* Macintosh 128K or Macintosh SE, its address space is limited
* to 2^24 = 16MiB. The RAM limits to 4MiB.
*
* With 512x342 1 bit per pixel screen, the size of the frame buffer
* is 0x5580 bytes.
*
* In Macintosh 128K [1], the frame buffer address is mapped from
* 0x1A700 to 0x1FC7F.
*
* In Macintosh SE [2], the frame buffer address is mapped from
* 0x3FA700 to 0x3FFC7F.
*
* The frame24_xxx memory banks mapping used the magic number to
* retrieve the offset. The memory write operation does twice:
* one for the guest OS and another for the host OS (the write operation
* above MacFrameBaseHost).
*
*
* See:
* [1] The Apple Macintosh Computer. http://www.1000bit.it/support/articoli/apple/mac128.pdf
* [2] Capturing Mac SE's video from PDS. http://synack.net/~bbraun/sevideo/
*
*/
void REGPARAM2 fram24_lput(uaecptr addr, uae_u32 l)
{
uaecptr page_off = addr & 0xffff;