physical address formula in 16 bit real mode: why multiply segment by 16?

Question

physical address=16*selector+offset
but i don't know why multiplying 16 by selector?

Answer 1

In order to be "programmer-compatible" with the Z80, yet still be able to use more than 64 kiB of memory, early Intel processors introduced memory segmentation. The 16-bit segment would be shifted left 4 bits (meaning multiplication by 16, and not 64 as your question claims) before being added to the 16-bit offset, resulting in a 20-bit address.

For programmers accustomed to the Z80, all that was required was to use the segments provided by the OS, and they would be able to use the given 64 kilobyte offset as they pleased. New programmers could do more sophisticated manipulation of the segments, allowing them to access 1 MiB of address space (the IBM PC cut it down to 640 kiB, but for their own reason).

Answer 2

The goal with the 8088/8086 was to provide a means by which a 16-bit processor could have a 1-megabyte addressing space, without programmers having to divide the address space into 64K chunks and worry about whether addresses crossed the boundaries between them. Though it has been much maligned, it is actually better than any other approach I've seen for working with addresses that exceed the register size. It often works relatively effortlessly in cases where data are naturally subdivided into items are 64K or less, and where it is acceptable to pad items to begin on 16-byte boundaries. The multiplication by 16 (instead of some larger or smaller number) was probably arbitrary, but it works out conveniently when listing things in hexadecimal notation, and it was a good compromise between using two small a number (which would have limited the address space to half a meg or less) or too large a number (which would have required more code to either deal with arbitrary offsets for allocated blocks, or else waste more memory padding blocks to a larger multiple).

The only major deficiencies in the 8088's segmentation design, really, are:

1. Two general-purpose segmentation registers aren't "quite" enough. A common pattern is "copy data from one object to another object, using a third object to translate it"; that pattern can only be implemented efficiently if one of the objects lives in either the code segment or on the stack. It's unfortunate this wasn't fixed until the 80386, where common practice was to ignore segments.
2. There wasn't any nice way to do any sort of arithmetic on segment registers. Instructions to add or subtract 0x1000 from DS or ES (or FS or GS), either if carry was set or unconditionally--eight (or sixteen) opcodes total--would have greatly facilitated segment handling.
3. There should have been "mov seg,immediate" instructions.

Even though programmers often grumbled about the 8088/8086 back in the day, it was far better than any other 16-bit processor I've seen then or since (nb: I consider the 68000 a 32-bit processor).

Answer 3

64? Sure? Read e.g. wikipedia. Multiplying by 64 is like shifting left by 6 bits (wikipedia says it should be 4, i.e. *16), i.e. is like saying the selector represents the most significant 16 bits of a 22 bits address (wikipedia reports 20). This is real mode as decribed in wikipedia too (better than I can do).

Answer 4

i don't know why multiplying 64 by selector?

A 16-bit pointer can easily address 64 KB.

The CPU designers wanted to be able to address 1 MB.

So instead of using a single 16-bit pointer, they specified that a pointer would be implemented by two registers, i.e. segment-plus-offset, where 'segment' is a 16-bit register whose value is multiplied by 16 in order to address the 1 MB.

The value (16) comes from dividing the desired address range (1 MB) by the natural addressibility of the 16-bit register size (64 KB) ... i.e. 16 come from 1 MB / 64 KB.