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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?
Mrstsn
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What is a segment used for?
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.
Justify that the virtual memory of 80386 is 64TB?
Max length of memory segment =4GB there r 8KB number of descriptor in GDT and 8KB in LDT . So total numbr of descriptors=16KB 4GB*16KB =64TB
What is the place holder in computer memory?
Computer memory usually refers to the RAM, there is key that you should know For each byte in RAM there is a separate address means if the RAM is 1 GB then its total addresses are 1*1024*1024 = 1048576 (0 - 1048575) addresses, Physically there are cells each can only contains 1 byte...... SO these CELL YOU CAN SAY IT THE PLACE HOLDER
What is minimum and maximum segment size in 8086?
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
What is the native heap size?
The heap refers to the free store, which basically means the total unused memory available to you. The physical amount of memory will vary from system to system and the amount of memory available will depend upon how much is currently in use. However memory fragmentation means that while there may be sufficient physical memory available to meet an allocation request it does not follow that there is a large enough block of contiguous memory available, resulting in an out of memory error. Modern systems use virtual memory addresses rather than physical ones. This allows the memory manager to physically move objects around in memory without affecting the virtual addresses allocated to those objects and thus makes it possible to consolidate memory fragments. Objects that are not in use can also be paged out to a disk file, thus making it seem like there's far more memory available than physically exists.
How segmentation and pipe lining in 8086 are related to each other?
The process of dividing total memory sizes to the segment of various sizes is called segmentation. The device which is used to fetch,decode and execute is called pipe lining.
What is the difference between physical memory and available memory?
Physical memory is how much total memory your computer actually has. Available memory is what memory you have that is not being used.
Why was segmentation originally introduced in 8088 architecture?
Memory segmentation is an attempt to address more memory than the processor architecture would normally allow.In the case of the 8086/8088, a 16 bit processor, you would normally expect addressibility of 64 kb, because that is what the instruction set is capable of developing as an effective address, either in the case of a direct address, an indirect address, or an indexed address, since all of its registers are 16 bits in size.What Intel did was provide four more registers called segment registers which would provide the base address of an address in physical memory to which the processor generated effective address would be added. The segment register is still 16 bits in size, but it is left shifted by four before being added to the effective address. This creates a physical address that is 20 bits in size, for a total address range of 1 mb.Note that you are still constrained to a segment size of 64 kb, in that you must stay within 64 kb unless you intend to change the value of one of the segment registers. This hampers the ability to access any arbitrary location in memory, effectively making it a two step operation - load the segment register - then access the offset address.In the 8086/8088 there are four segment registers; Code Segment (CS), Data Segment (DS), Stack Segment (SS), and Extra Segment (ES). All opcode access is from CS. Default data access is from DS, unless a segment prefix is applied. All stack operations are from SS. Certain repeated string operations take place between DS and ES.Because of the segmented architecture, the concept of near and far grew up with the original PC and DOS and Windows. Basically, a near address was a 16 bit address that assumed the current segment, while a far address was a 32 bit address that contained both a segment and an offset. Note that the concept of a flat 32 bit address did not come into full play until true 32 bit operating systems hit the street, and that did not occur until the introduction of the 80386.
What is minimum size of a segment in Intel 8086 Why?
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient. In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment size is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
How many IPv6 address are possible?
In theory there are 264 available network allocations in ipV6. It is not known how many are actually in use at this time. For a very clear comparison, in IPv4 there is a total of 4,294,967,296 IP addresses. With IPv6, there is a total of 18,446,744,073,709,551,616 IP addresses in a single /64 allocation.
How do you find if processor is 64 bit or 32 bit?
A 32-bit system has a maximum 2^32 unique memory addresses, which is 4,294,967,296 addresses in total. Each address refers to a byte (the smallest unit of storage), thus this allows a maximum address space of 4 gigabytes. Each additional bit doubles the number of available addresses, thus a 64-bit system supports a maximum of 18,446,744,073,709,551,616 bytes, which is more memory than exists on the planet. Note that the length of a byte is system-defined.
What do you call the total loss of memory?
amneisa
