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The 8086/8088 is a 16 bit processor running on a 16 bit (8086) or 8 bit (8088) bus with a 20 bit address bus. In order to obtain the extra 4 bits of addressibility, Intel designed segment registers that are effectively multiplied by four and then added to the 16 bit offset address generated by the instruction. This yields 64K segments of 64KB each, although they overlap each other at a distance of 16 bytes.
Memory segmentation in the 8086/8088 is used to allow the processor to access more than 64kb of memory, even though it is only a 16-bit processor. Each segment register allows access to one of 64k 64kb segments, each overlapping by 16 bytes, with the total addressibility being 1mb.
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When does segmentation fault occur for a prog?
You either reference memory that is non existent, or you attempt to modify memory that is read only. This is usually a result of failure to properly initialize or use pointers or arrays.
What is use of offset address in 8086 mp?
Offset address is also known as displacement.By adding this offset value to a base address,address of a specific locaction in memory can be accessed
What is the working of 8087 NDP microprocessor?
The 8087 Numeric Data Processor is an adjunct to the 8086/8088 microprocessor, that gives the 8086/8088 floating point capability and 8 more registers. The 8087 integrates itself with the 8086/8088 in such a way that the pair actually becomes one processor, appearing to have the extra instructions and registers to start with.Since it was an optional add-on to the system, many run-time libraries would detect whether or not the 8087 was present, and either use it or emulate it.This separation of functionality into two chips was maintained in the product line up until the 80386DX and 80486, although the NDP was called something else (such as the 80287, and 80387) at which point the NDP became a permanent part of the instruction set.
What is the use of insertion queue in 8086?
Maybe you mean the prefetch queue?
Advantages of microprocessor based system?
There are many different advantages of a microprocessor based system. Some of these include that it only uses 3 components so it is simple to use and install and it will increase your memory.
What is memory interfacing in 8086 microprocessor?
The 8086/8088 is a 16 bit computer running on a 20 bit address bus. Processes use a segmented memory architecture to access one of four 64kb memory segments from a physical space of 1mb.
How the speed of 8086 microprocessor can be increased?
To increase the speed of the 8086, you need to increase the clock speed, reduce the number of wait states, or both. You could also optimize your code so that it runs faster. Since the 8086 is a segmented memory architecture, it is more efficient to use operands in one segment and to make near references to them.
Is there any on board computer with microprocessor 8086 that use to measure temperature and pressure?
Yes, there are on-board computers with microprocessor 8086 that can be used to measure temperature and pressure. These computers can be equipped with sensors for temperature and pressure measurements, and programmed to process and display the data collected from these sensors in real-time. The 8086 microprocessor is capable of handling the data processing requirements for these measurements.
Why in memory segmentation of 8086 it use 64Kb instead of 1Mb?
Each segment in the 8086/8088 is 64KB because that is how Intel designed the microprocessor. The offset address is 16 bits, making the allowable range 64KB. See the related questions link for a further discussion of segmented architecture.
How many pins are present in 8085 and 8086 microprocessor?
There are 256 ports available in the 8085 microprocessor. The IN and OUT instructions have an 8-bit port number, and that is where the 256 comes from.In order to use ports, the hardware addressing system must decode IO/M-. Some implementations don't do this, so they map IO addresses to memory addresses. In that case, you could say there are 65536 possible IO addresses, but that is not the same as ports, because ports are specific to the IN and OUT instructions.The other "problem" with IN and OUT is that you cannot specify the address in a register, while you can do so with indirect memory addressing.
What is the use of debug in 8086 microprocessor?
The debug flag in the 8086/8088 microprocessor causes one instruction to be executed, followed immediately by a debug interrupt. The flag then gets turned off, so you get get nested debug interrupts. This is used by a debugger program to single step a process that it is debugging (the debugee, so to speak).
What is the use of Intel 8257 microprocessor?
Its a Programmable Direct Memory Access (DMA) controller.
How do you access 1 GB memory with 8085?
You need some kind of memory expander, which maps a frame of addresses to a location in physical memory. Better, use an 8086/8088.
What is memory space in microprocessor?
memory space in microprocessor means cache ,it is the part of the microprocessor which contains the memory to store instructions which are used to perform different functions by the processor. where cache1 is referred to memory in microprocessor and cache2 is placed on motherboard which also contains memory to store instructions. Memory space can also mean the total size of virtual memory that a CPU can address, and the layout (flat or segmented) of this space. Most modern CPUs found in PCs (AMD and Intel x86_64 chips) are 64-bit CPUs, but, due to cost and practical use considerations, limit their virtual memory space to be 48-bits (or less).
How is protection affected in sgmentation of memory?
Protection in the segmentation of memory is protected by disallowing a process that is in use to access the memory of other processes that are not being used. It is a barrier between processes to protectÊthose not in use from possible virus infection or malicious software infection.
Which microprocessor did sputnik use?
Sputnik did not use a microprocessor.
When does segmentation fault occur for a prog?
You either reference memory that is non existent, or you attempt to modify memory that is read only. This is usually a result of failure to properly initialize or use pointers or arrays.
