8086 CPU ARCHITECTURE
The microprocessors functions as the CPU in the stored program model of the digital computer. Its job is to generate all system timing signals and synchronize the transfer of data between memory, I/O, and itself. It accomplishes this task via the three-bus system architecture previously discussed.
The microprocessor also has a S/W function. It must recognize, decode, and execute program instructions fetched from the me  mory unit. This requires an Arithmetic-Logic Unit (ALU) within the CPU to perform arithmetic and logical (AND, OR, NOT, compare, etc) functions.
The 8086 CPU is organized as two separate processors, called the Bus Interface Unit (BIU) and the Execution Unit (EU). The BIU provides H/W functions, including generation of the memory and I/O addresses for the transfer of data between the outside world -outside the CPU, that is- and the EU.
The EU receives program instruction codes and data from the BIU, executes these instructions, and store the results in the general registers. By passing the data back to the BIU, data can also be stored in a memory location or written to an output device. Note that the EU has no connection to the system buses. It receives and outputs all its data thru the BIU.
The only difference between an 8088 microprocessor and an 8086 microprocessor is the BIU. In the 8088, the BIU data bus path is 8 bits wide versus the 8086's 16-bit data bus. Another difference is that the 8088 instruction queue is four bytes long instead of six.
The important point to note, however, is that because the EU is the same for each processor, the programming instructions are exactly the same for each. Programs written for the 8086 can be run on the 8088 without any changes.
Although the 8086/88 still functions as a stored program computer, organization of the CPU into a separate BIU and EU allows the fetch and execute cycles to overlap. To see this, consider what happens when the 8086 or 8088 is first started.
1. The BIU outputs the contents of the instruction pointer register (IP) onto the address bus, causing the selected byte or word to be read into the BIU.
2. Register IP is incremented by 1 to prepare for the next instruction fetch.
3. Once inside the BIU, the instruction is passed to the queue. This is a first-in, first-out storage register sometimes likened to a "pipeline".
4. Assuming that the queue is initially empty, the EU immediately draws this instruction from the queue and begins execution.
5. While the EU is executing this instruction, the BIU proceeds to fetch a new instruction. Depending on the execution time of the first instruction, the BIU may fill the queue with several new instructions before the EU is ready to draw its next instruction.
The BIU is programmed to fetch a new instruction whenever the queue has room for one (with the 8088) or two (with the 8086) additional bytes. The advantage of this pipelined architecture is that the EU can execute instructions almost continually instead of having to wait for the BIU to fetch a new instruction.
There are three conditions that will cause the EU to enter a "wait" mode. The first occurs when an instruction requires access to a memory location not in the queue. The BIU must suspend fetching instructions and output the address of this memory location. After waiting for the memory access, the EU can resume executing instruction codes from the queue (and the BIU can resume filling the queue).
One other condition can cause the BIU to suspend fetching instructions. This occurs during execution of instructions that are slow to execute. For example, the instruction AAM (ASCII Adjust for Multiplication) requires 83 clock cycles to complete. At four cycles per instruction fetch, the queue will be completely filled during the execution of this single instruction. The BIU will thus have to wait for the EU to pull over one or two bytes from the queue before resuming the fetch cycle.
A subtle advantage to the pipelined architecture should be mentioned. Because the next several instructions are usually in the queue, the BIU can access memory at a somewhat "leisurely" pace. This means that slow-mem parts can be used without affecting overall system performance.
PROGRAMING MODEL
As a programmer of the 8086 or 8088 you must become familiar with the various registers in the EU and BIU.
The data group consists of the accumulator and the BX, CX, and DX registers. Note that each can be accessed as a byte or a word. Thus BX refers to the 16-bit base register but BH refers only to the higher 8 bits of this register. The data registers are normally used for storing temporary results that will be acted on by subsequent instructions.
The pointer and index group are all 16-bit registers (you cannot access the low or high bytes alone). These registers are used as memory pointers. Sometimes a pointer reg will be interpreted as pointing to a memory byte and at other times a memory word. As you will see, the 8086/88 always stores words with the high-order byte in the high-order word address.
Register IP could be considered in the previous group, but this register has only one function -to point to the next instruction to be fetched to the BIU. Register IP is physically part of the BIU and not under direct control of the programmer as are the other pointer registers.
Six of the flags are status indicators, reflecting properties of the result of the last arithmetic or logical instructions. The 8086/88 has several instructions that can be used to transfer program control to a new memory location based on the state of the flags.
Three of the flags can be set or reset directly by the programmer and are used to control the operation of the processor. These are TF, IF, and DF.
The final group of registers is called the segment group. These registers are used by the BIU to determine the memory address output by the CPU when it is reading or writing from the memory unit. To fully understand these registers, we must first study the way the 8086/88 divides its memory into segments.
SEGMENTED MEMORY
Even though the 8086 is considered a 16-bit processor, (it has a 16-bit data bus width) its memory is still thought of in bytes. At first this might seem a disadvantage:
Why saddle a 16-bit microprocessor with an 8-bit memory?
Actually, there are a couple of good reasons. First, it allows the processor to work on bytes as well as words. This is especially important with I/O devices such as printers, terminals, and modems, all of which are designed to transfer ASCII-encoded (7- or 8-bit) data.
Second, many of the 8086's (and 8088's) operation codes are single bytes. Other instructions may require anywhere from two to seven bytes. By being able to access individual bytes, these odd-length instructions can be handled.
We have already seen that the 8086/88 has a 20-bit address bus, allowing it to output 210, or 1'048.576, different memory addresses. As you can see, 524.288 words can also be visualized.
As mentioned, the 8086 reads 16 bits from memory by simultaneously reading an odd-addressed byte and an even-addressed byte. For this reason the 8086 organizes its memory into an even-addressed bank and an odd-addressed bank.
With regard to this, you might wonder if all words must begin at an even address. Well, the answer is yes. However, there is a penalty to be paid. The CPU must perform two memory read cycles: one to fetch the low-order byte and a second to fetch the high-order byte. This slows down the processor but is transparent to the programmer.
The last few paragraphs apply only to the 8086. The 8088 with its 8-bit data bus interfaces to the 1 MB of memory as a single bank. When it is necessary to access a word (whether on an even- or an odd-addressed boundary) two memory read (or write) cycles are performed. In effect, the 8088 pays a performance penalty with every word access. Fortunately for the programmer, except for the slightly slower performance of the 8088, there is no difference between the two processors.
MEMORY MAP
Still another view of the 8086/88 memory space could be as 16 64K-byte blocks beginning at hex address 000000h and ending at address 0FFFFFh. This division into 64K-byte blocks is an arbitrary but convenient choice. This is because the most significant hex digit increments by 1 with each additional block. That is, address 20000h is 65.536 bytes higher in memory than address 10000h. Be sure to note that five hex digits are required to represent a memory address.
The diagram is called a memory map. This is because, like a road map, it is a guide showing how the system memory is allocated. This type of information is vital to the programmer, who must know exactly where his or her programs can be safely loaded.
Note that some memory locations are marked reserved and others dedicated. The dedicated locations are used for processing specific system interrupts and the reset function. Intel has also reserved several locations for future H/W and S/W products. If you make use of these memory locations, you risk incompatibility with these future products.
SEGMENT REGISTERS
Within the 1 MB of memory space the 8086/88 defines four 64K-byte memory blocks called the code segment, stack segment, data segment, and extra segment. Each of these blocks of memory is used differently by the processor.
The code segment holds the program instruction codes. The data segment stores data for the program. The extra segment is an extra data segment (often used for shared data). The stack segment is used to store interrupt and subroutine return addresses.
You should realize that the concept of the segmented memory is a unique one. Older-generation microprocessors such as the 8-bit 8086 or Z-80 could access only one 64K-byte segment. This mean that the programs instruction, data and subroutine stack all had to share the same memory. This limited the amount of memory available for the program itself and led to disaster if the stack should happen to overwrite the data or program areas.
The four segment registers (CS, DS, ES, and SS) are used to "point" at location 0 (the base address) of each segment. This is a little "tricky" because the segment registers are only 16 bits wide, but the memory address is 20 bits wide. The BIU takes care of this problem by appending four 0's to the low-order bits of the segment register. In effect, this multiplies the segment register contents by 16.
The point to note is that the beginning segment address is not arbitrary -it must begin at an address divisible by 16. Another way if saying this is that the low-order hex digit must be 0.
Also note that the four segments need not be defined separately. Indeed, it is allowable for all four segments to completely overlap (CS = DS = ES = SS).
Memory locations not defined to be within one of the current segments cannot be accessed by the 8086/88 without first redefining one of the segment registers to include that location. Thus at any given instant a maximum of 256 K (64K * 4) bytes of memory can be utilized. As we will see, the contents of the segment registers can only be specified via S/W. As you might imagine, instructions to load these registers should be among the first given in any 8086/88 program.
LOGICAL AND PHYSICAL ADDRESS
Addresses within a segment can range from address 00000h to address 0FFFFh. This corresponds to the 64K-byte length of the segment. An address within a segment is called an offset or logical address. A logical address gives the displacement from the address base of the segment to the desired location within it, as opposed to its "real" address, which maps directly anywhere into the 1 MB memory space. This "real" address is called the physical address.
What is the difference between the physical and the logical address?
The physical address is 20 bits long and corresponds to the actual binary code output by the BIU on the address bus lines. The logical address is an offset from location 0 of a given segment.
When two segments overlap it is certainly possible for two different logical addresses to map to the same physical address. This can have disastrous results when the data begins to overwrite the subroutine stack area, or vice versa. For this reason you must be very careful when segments are allowed to overlap.
You should also be careful when writing addresses on paper to do so clearly. To specify the logical address XXXX in the stack segment, use the convention SS:XXXX, which is equal to [SS] * 16 + XXXX.
ADVANTAGES OF SEGMENTED MEMORY
Segmented memory can seem confusing at first. What you must remember is that the program op-codes will be fetched from the code segment, while program data variables will be stored in the data and extra segments. Stack operations use registers BP or SP and the stack segment. As we begin writing programs the consequences of these definitions will become clearer.
An immediate advantage of having separate data and code segments is that one program can work on several different sets of data. This is done by reloading register DS to point to the new data. Perhaps the greatest advantage of segmented memory is that programs that reference logical addresses only can be loaded and run anywhere in memory. This is because the logical addresses always range from 00000h to 0FFFFh, independent of the code segment base. Such programs are said to be relocatable, meaning that they will run at any location in memory. The requirements for writing relocatable programs are that no references be made to physical addresses, and no changes to the segment registers are allowed.
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