Intel 8086 and 8088

Address multiplexing allows you to use fewer pins on the processor, and thus fewer bus lines. So instead of having some bus lines for the address, and some more for the data, you put the address on the data line, it gets read, then you put the data on the same lines, and it gets read and stored at the previously read address. For the 8085, it allowed the design to add one pin, but cut 8, for a net gain (loss?) of 7 pins (reduced physical/manufacturing complexity at the cost of increased logical/programming complexity).

Computers have different devices operating at different speeds. As such, there are often multiple devices competing for the bus at the same time. To allow transactions to occur in parallel instead of "taking a ticket", the system needs to be able to hold data when it becomes available but the bus is busy, until the bus gets freed. It holds that data in a buffer.

I hope that was clear enough. If not, feel free to ask for clarification on anything you didn't understand

Implicit addressing modes are of the assumption that the data is in predefined registers. also Known as Zero address instructions:

Eg: XLAT ; assumes the operands in AX and BX

AAM ;operates on the contents of AX only

ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.

Tristate devices are used in bus based systems to allow multiple bus drivers to control the bus, each at different times, while all the rest are allowed to read the bus.

Only one device can drive the bus at any one time. All the others "tristate" or float, so they neither drive the bus low nor high.

mov

The 8086 has a 16 bit bus, while the 8088 has an 8 bit bus. In both A0 through A7 are shared as D0 through D7. The the 8086, A8 through A15 is also shared as D8 through D15.

External record is simply defined as case of data that is collected from outside. This is commonly used as a representation of proof in form of external files.

A two-byte instruction gives the specific function instruction in two bytes, or two words. The first specifies the opcode, which tells the microprocessor what operation will occur. The second specifies the operand, or the data that the operation is done on.

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Give a brief description of the type of employment or position you desire.Give a brief description of the type of employment or position you desire.

Which processor has a 32-bit Data Bus and a 24-bit Address Bus?

The ORA A instruction does not change the contents of the accumulator. It does, however, set flags to indicate something about the value of the accumulator. In particular, Z means the result is zero, N means it is negative, and P means it has an odd number of ones.

10. 210 = 1024.

A 7 segment display is an I/O device, and it is not necessarily used in microprocessors. The choice of I/O devices is a function of system design, not microprocessor design.

MCA

A) Segment: 0000: Range: 00000-0FFFF

B) Segment: FFFF: Ranges: FFFF0-FFFFF and 00000-0FFEF

Yes.

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The 8086 takes two cycles to fetch words from an odd address because the Bus Interface Unit (BIU) actually only deals with words, which are two bytes at an even address. If you perform an operation on a byte, the BIU will do a word operation, but it will leave the other byte alone. If you perform a word operation at an even address, the BIU can do the operation in one cycle. If the word, however, is at an odd address, the BIU must use two cycles, leaving the respective "other" byte alone in each case. For performance sake, always align your data with the bus design.

I asked what is the application of bus architecture

16 bit addition

Load effective address, in assembler terms, and in the case of for example "lea eax, [esp-8]" means "please compute and put the address of [esp-8] into the register eax".

In other terms, "mov" works only on values, except you tell it to get the address of your source arg (you could do mov eax, offset [ebp-8]), while "lea" works on addresses only.

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.

Physical servers (also referred to as boxes) are represented by physically or virtually separate computers with their own operating systems (Linux, FreeBSD Unix, or Windows).

protected mode