Xref: utzoo comp.sys.ibm.pc.misc:7963 comp.sys.ibm.pc.hardware:7004 comp.os.msdos.programmer:4312 Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!batcomputer!cornell!rochester!pt.cs.cmu.edu!o.gp.cs.cmu.edu!ralf From: ralf+@cs.cmu.edu (Ralf Brown) Newsgroups: comp.sys.ibm.pc.misc,comp.sys.ibm.pc.hardware,comp.os.msdos.programmer Subject: Re: Memory above 640K being redundant (or something like that) Message-ID: <1991Mar29.011437.15906@cs.cmu.edu> Date: 29 Mar 91 01:14:37 GMT References: <1019@stewart.UUCP> Sender: netnews@cs.cmu.edu (USENET News Group Software) Organization: School of Computer Science, Carnegie Mellon Lines: 31 In article <1019@stewart.UUCP> jerry@stewart.UUCP (Jerry Shekhel) writes: }ralf+@cs.cmu.edu (Ralf Brown) writes: }>>Ok, so WHAT address does a 286/386/486 jump to when it resets, (or at }>>boot time)?? }> }>286 = FFFFFF:0 }>386/486 = FFFFFFFF:0 }> }>In other words, the start of the highest segment in the address space. } }What!? I always thought that both the 286 and the 386/486 are in real mode }when they reset, so they still jump to FFFF:0. Am I wrong? How can the }286 (or 386/486) jump to an address above 1MB if the protected mode selectors }for the segment registers haven't been created yet? If the 386/486 supports }4GB segments, what does "highest segment in the address space" mean? Well, I've double-checked my copies of the 286 and 486 PRMs, and found that I was slightly off: 286 = FF000:FFF0 = FFFFF0h absolute 486 = FFFF0000:0000FFF0 = FFFFFFF0h absolute (in both cases, 16 bytes before the top of the address space). As to how this is possible in the real mode in which the processor starts, the CS register is set to F000h, but the invisible descriptor cache for CS holds the full address. After the first far jump, the high four (twelve) bits of the descriptor cache address field are forced to zero. -- {backbone}!cs.cmu.edu!ralf ARPA: RALF@CS.CMU.EDU FIDO: Ralf Brown 1:129/3.1 BITnet: RALF%CS.CMU.EDU@CMUCCVMA AT&Tnet: (412)268-3053 (school) FAX: ask DISCLAIMER? Did | It isn't what we don't know that gives us trouble, it's I claim something?| what we know that ain't so. --Will Rogers