CODEWITHSUNDEEP
memory-managementassemblyx86
Ali
Ali

Reputation: 1286

Why do segments begin on paragraph boundaries?

In real mode segmented memory model, a segment always begins on a paragraph boundary. A paragraph is 16 bytes in size so the segment address is always divisible by 16. What is the reason/benefit of having segments on paragraph boundaries?

Upvotes: 4

Views: 4162

Answers (3)

lonewolf
lonewolf

Reputation: 69

The Segment register stores the address of the memory location where that segment starts. But Segment registers stores 16 bit information. This 16 bit is converted to 20 bits by appending 4 bits of 0 to the right end of the address. If a segment register contains 1000H then it is left shifted to get 10000H. Now it is 20 bits.

While converting we added 4 bits of 0 to the end of the address. So every segment in memory must begin with memory location where the last 4 bits are 0.

For ex:

If a segment starts at 10001H memory location, we cannot access it because the last 4 bits are not 0. Any address in segment register will be appended with 4 bits in right end to convert to 20bits. So there is no way of accessing such an address.

Upvotes: 1

Michael Burr
Michael Burr

Reputation: 340198

It's not so much a benefit as an axiom - the 8086's real mode segmentation model is designed at the hardware level such that a segment register specifies a paragraph boundary.

The segment register specified the base address of the upper 16 bits of the 8086's 20 bit address space, the lower 4 bits of that base address were in essence forced to zero.

The segmented architecture was one way to have the 8086's 16-bit register architecture be able to address a full megabyte(!) of address space (which requires 20 bits of addressing).

For a little more history, the next step that Intel took in the x86 architecture was to abstract the segment registers from directly defining the base address. That was the 286 protected mode, where the segment register held a 'selector' that instead of defining the bits for a physical base address was an index into a a set of descriptor tables that held information about the physical address, permissions for access to the physical memory, and other things.

The segment registers in modern 32-bit or larger x86 processors still do that. But with the address offsets being able to specify a full 32-bits of addressing (or 64-bits on the x64 processors) and page tables able to provide virtual memory semantics within the segment defined by a selector, the programming model essentially does away with having to manipulate segment registers at the application level. For the most part, the OS sets the segment registers up once, and nothing else needs to deal with them. So programmers generally don't even need to know they exist anymore.

Upvotes: 4

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798576

The 8086 had 20 address lines. The segment was mapped to the top 16, leaving an offset of the bottom 4 lines, or 16 addresses.

Upvotes: 3

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