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Write a program using the following figure that determines and displays the physical addresses and the accessible physical addresses for each of the above logical addresses?
The Physical address can be calculated asPhysical address = Base address + Offset.So take Base,Offsets,Lengths in to different arrays and check whether offset is less than length , if it is then the physical address would be base+ Offset , if not then that is a segment error.
How the Memory Management Unit converts the logical addresses to physical addresses?
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.
How does memory segmentation increase processors speed?
The segmentation function maintains a segment table that includes physical addresses of the segment, size, and other data. Segmentation speeds up a computer's information retrieval by assigning related data into a “segment table” between the CPU and the physical memory.
As a memory segment is 64k does it mean that 64k are the total addresses in a segment if 64K is the total addresses in a segment then what is the sizebytes of one segment 64k?
Yes, if a memory segment is 64K, it means there are 64K total addresses in that segment. Since 1K (kilobyte) equals 1024 bytes, a 64K segment would have a total size of 64 * 1024 bytes, which equals 65,536 bytes. Therefore, the size of one segment is 65,536 bytes.
What is overlapping segment addresses?
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.
How do you know the starting and ending address of E000H in segment register?
The starting and ending addresses of segment E000H is E0000H and EFFFFH.
Why a segment can not start from the physical address 77777?
For any segment base address, segment first physical address will have 0 in the least significant position in hexadecimal format. Let say, our Segment base =0x1234, and we calculate segment first physical address as 0x12340
How can physical address be calculated using protected mode?
In protected mode, physical addresses are calculated using a combination of segment selectors and offsets. A segment selector points to a descriptor in the Global Descriptor Table (GDT) or Local Descriptor Table (LDT), which contains the base address of the segment along with its limit and access rights. To compute the physical address, the base address from the descriptor is added to the offset provided by the program, resulting in the final physical address used by the CPU. This segmentation allows for better memory management and protection in multitasking environments.
Is a segment an idea and not a physical object?
it is a idea
What is segment overlapping?
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.
Which program segment at run time is used to store interrupt and subroutine return addresses?
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.
Is RAM a linear or nonlinear data structure?
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.
