Each segment register is multiplied by 16 before being added to the effective address during physical address computation. As a result, sequential values of segment registers result in physical addresses that are 16 bytes apart, yet, the maximum distance, based on effective address is 64 KB. That is why the segmented memory addressing approach in the 8086/8088 is overlapping - there are 64 K different combinations of segment register / effective address pair that will result is the same physical address. Well - sort of - its not wrap around - its a sliding window - call it a 16 byte paragraph size with an addressability range of 64 KB.
A segment is a chunk (segment) of memory that is 64Kb in size. Due to the design of the 8086/8088 there are 64K possible segments, ecah overlapping the next by 16 bytes, for a total addressibility of 1 Mb. In the instruction model, a segment is the locus of addresses that can be reached in one instruction, without stopping to load a new value into a segment register. It is also called a near, or 16 bit address.
Each segment register is multiplied by 16 before being added to the effective address during physical address computation. As a result, sequential values of segment registers result in physical addresses that are 16 bytes apart, yet, the maximum distance, based on effective address is 64 KB. That is why the segmented memory addressing approach in the 8086/8088 is overlapping - there are 64 K different combinations of segment register / effective address pair that will result is the same physical address. Well - sort of - its not wrap around - its a sliding window - call it a 16 byte paragraph size with an addressability range of 64 KB.
The starting and ending addresses of segment E000H is E0000H and EFFFFH.
In the 8086/8088, the logical address corresponds to a segment register, such as CS (Code Segment), DS (Data Segment), SS (Stack Segment) and ES (Extra Segment). The segment register is selected by context, or it is explicitly selected using a segment override prefix. The segment register is left shifted 4 bits into a 20-bit temporary register. This is the same as multiplying it by 16. Then the logical address is added to that result. The final result is the physical address.
The addresses for the 8086 range from 00000 hex to FFFFF hex, but they are accessed as offsets within overlapping 64-kB blocks.
If this is a homework assignment, you really should try to answer it on your own first, otherwise the value of the reinforcement of the lesson due to actually doing the assignment will be lost on you.Interrupt and subroutine return addresses are stored on the stack, so the stack segment is used for interupt and subroutine addresses.
That depends on the computer architecture. Usually, we treat RAM as a linear data structure, with all addresses in one linear address space. But on some architectures RAM is segmented, meaning addresses have both a segment and an address within a segment.
A) Segment: 0000: Range: 00000-0FFFF B) Segment: FFFF: Ranges: FFFF0-FFFFF and 00000-0FFEF
They are overlapping events.They are overlapping events.They are overlapping events.They are overlapping events.
Tort law is the segment of law that addresses cases involving civil wrongs. A tort is simply an injury.
In a pure segmentation architecture, segments are allocated like variable partitions, although the memory management hardware is involved in decoding addresses. Pure segmentation addresses replace the page identifier in the virtual address with a segment identifier, and find the proper segment (not page) to which to apply the offset.
it reduces internal fragmentation....on 8086 its limited to a max of just 16b per segment of each process....in other words if they weren't overlapping this would be 2^16! btw I'd like to hear more possible explanations too!