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DJSAN10 28 Posting Whiz in Training

Was there any specific reason IBM decided to start loading BIOS into memory from 640k rather than from the beginning( 0th byte) ?I tried searching on the Internet but did not get a satisfactory answer.

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Short answer: IBM did not put the system ROM at physical address 0 — it was mapped into the upper 384 KB of the original 1 MB PC address space (the area above 640 KB). The important point is that the 8088/8086 used in the IBM PC starts execution at a physical reset vector near the top of that 1 MB space, so the BIOS had to be visible there rather than at address 0. See the note on conventional memory and the PC memory map for background. (Conventional memory).

Why that top-of-memory location? On the 8088/8086 the CPU reset entry point is at physical address 0xFFFF0 (16 bytes below 1 MB), so startup code must be reachable at that address. Placing the ROM at the top let the CPU fetch the reset vector without extra hardware remapping. It also left the low 0–640 KB region free for RAM, while the 640 KB–1 MB gap could hold video memory, video BIOS and other adapter option ROMs (C0000h, D0000h, etc.). That layout became the de facto standard and is described in historical PC technical references and common boot docs. (BIOS).

Practical consequences and why the design stuck: device manufacturers and DOS-era software expected the upper region for ROMs and MMIO, so compatibility required keeping the BIOS/option-ROM mappings stable. Later machines often “shadowed” the ROM contents into RAM at the same addresses for speed, but the original physical mapping persisted for compatibility.

Reference note for : the confusion usually comes from mixing “start of the upper memory area (640 KB)” with the BIOS’s actual location near 0xF0000. The CPU reset vector and the need to reserve space for video/adapter ROMs explain why IBM chose the upper portion of the 1 MB range rather than the 0th byte.

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