1 / 34

Computers organization & Assembly Language

Computers organization & Assembly Language. Chapter 1 THE 80x86 MICROPROCESSOR. Registers inside 8088/8086 , IA-32 Instruction Execution Cycle IA-32 Memory Management. Inside the 8088/8086. Registers inside 8088/8086. Registers are used to store information temporarily.

emiko
Download Presentation

Computers organization & Assembly Language

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Computers organization& Assembly Language Chapter 1 THE 80x86 MICROPROCESSOR

  2. Registers inside 8088/8086 , IA-32 • Instruction Execution Cycle • IA-32 Memory Management

  3. Inside the 8088/8086

  4. Registers inside 8088/8086 • Registers are used to store information temporarily. • That information could be one or two bytes of data to be processed or the address of data. • The general-purpose registers in 8088/86 microprocessors can be accessed as either 16-bit or 8-bit registers. • All other registers can be accessed only as the full 16 bits. • In the 8088/86. data types are either 8 or 16 bits. To access 12-bit data. for example, a 16-bit register must be used with the highest 4 bits set to 0.

  5. Registers inside 8088/8086 • In the 8088/86. data types are either 8 or 16 bits. • To access 12-bit data. for example, a 16-bit register must be used with the highest 4 bits set to 0.

  6. Registers inside 8088/8086

  7. 32-bit General-Purpose Registers EAX EBP EBX ESP ECX ESI EDX EDI 16-bit Segment Registers EFLAGS CS ES SS FS EIP DS GS Basic Program Execution Registers IA-32 • Registers are high speed memory inside the CPU • Eight 32-bit general-purpose registers • Six 16-bit segment registers • Processor Status Flags (EFLAGS) and Instruction Pointer (EIP)

  8. General-Purpose Registers • Used primarily for arithmetic and data movement • moveax, 10 move constant 10 into register eax • Specialized uses of Registers • EAX – Accumulator register • Automatically used by multiplication and division instructions • ECX – Counter register • Automatically used by LOOP instructions • ESP – Stack Pointer register • Used by PUSH and POP instructions, points to top of stack • ESI and EDI – Source Index and Destination Index register • Used by string instructions • EBP – Base Pointer register • Used to reference parameters and local variables on the stack

  9. Accessing Parts of Registers • EAX, EBX, ECX, and EDX are 32-bit Extended registers • Programmers can access their 16-bit and 8-bit parts • Lower 16-bit of EAX is named AX • AX is further divided into • AL = lower 8 bits • AH = upper 8 bits • ESI, EDI, EBP, ESP have only 16-bit names for lower half

  10. Accessing Parts of Registers

  11. Special-Purpose & Segment Registers • EIP = Extended Instruction Pointer • Contains address of next instruction to be executed • EFLAGS = Extended Flags Register • Contains status and control flags • Each flag is a single binary bit • Six 16-bit Segment Registers • Support segmented memory • Six segments accessible at a time • Segments contain distinct contents • Code • Data • Stack

  12. Floating-Point, MMX, XMM Registers • Floating-point unit performs high speed FP operations • Eight 80-bit floating-point data registers • ST(0), ST(1), . . . , ST(7) • Arranged as a stack • Used for floating-point arithmetic • Eight 64-bit MMX registers • Used with MMX instructions • Eight 128-bit XMM registers • Used with SSE instructions

  13. Next ... • IA-32 Registers • Instruction Execution Cycle • IA-32 Memory Management

  14. Fetch-Execute Cycle • Each machine language instruction is first fetched from the memory and stored in an Instruction Register(IR). • The address of the instruction to be fetched is stored in a register called Program Counter or simply PC. In some computers this register is called the Instruction Pointer or IP. • After the instruction is fetched, the PC (or IP) is incremented to point to the address of the next instruction. • The fetched instruction is decoded (to determine what needs to be done) and executed by the CPU.

  15. Computers Organization & Assembly Language Clock • synchronizes all CPU and BUS operations • machine (clock) cycle measures time of a single operation • clock is used to trigger events

  16. Instruction Fetch Obtain instruction from program storage Instruction Decode Determine required actions and instruction size Operand Fetch Infinite Cycle Locate and obtain operand data Execute Compute result value and status Writeback Result Deposit results in storage for later use Instruction Execute Cycle

  17. PC program I1 I2 I3 I4 . . . fetch memory read op1 op2 registers registers instruction I1 register decode write write ALU flags execute (output) Instruction Execution Cycle – cont'd • Instruction Fetch • Instruction Decode • Operand Fetch • Execute • Result Writeback

  18. Pipelining • There are two ways to make the CPU process information faster: • increase the working frequency • or change the internal architecture of the CPU. (Pipelining) • Pipelining is allow the CPU to fetch and execute at the same time. • Pipelining in the 8088/86 by splitting the internal structure of the microprocessor into two sections: the execution unit (EU) and the bus interface unit (BIU). • These two sections work simultaneously. The BIU accesses memory and peripherals while the EU executes instructions previously fetched. • This works only if the BIU keeps ahead of the EU; thus the BIU of the 8088/86 has a buffer, or queue. • The buffer is 4 bytes long in the 8088 and 6 bytes in the 8086. • If any instruction takes too long to execute, the queue is filled to its maximum capacity and the buses will sit idle. • The BIU fetches a new instruction whenever the queue has room for 2 bytes in the 6-byte 8086 queue, and for 1 byte in the 4-byte 8088 queue.

  19. Pipelined vs. Nonpipelined

  20. Stages S1 S2 S3 S4 S5 S6 1 I-1 2 I-1 3 I-1 4 I-1 5 I-1 Cycles 6 I-1 7 I-2 8 I-2 9 I-2 10 I-2 11 I-2 12 I-2 Pipelined Execution • Instruction execution can be divided into stages • Pipelining makes it possible to start an instruction before completing the execution of previous one For k stages and n instructions, the number of required cycles is: k + n – 1 Non-pipelined execution Wasted clock cycles Pipelined Execution

  21. Stages exe S1 S2 S3 S4 S5 S6 1 I-1 2 I-2 I-1 3 I-3 I-2 I-1 4 I-3 I-2 I-1 Cycles 5 I-3 I-1 6 I-2 I-1 7 I-2 I-1 8 I-3 I-2 9 I-3 I-2 10 I-3 11 I-3 Wasted Cycles (pipelined) • When one of the stages requires two or more clock cycles to complete, clock cycles are again wasted • Assume that stage S4 is the execute stage • Assume also that S4 requires 2 clock cycles to complete • As more instructions enter the pipeline, wasted cycles occur • For k stages, where one stage requires 2 cycles, n instructions require k + 2n – 1 cycles

  22. Superscalar Architecture • A superscalar processor has multiple execution pipelines • The Pentium processor has two execution pipelines • Called U and V pipes • In the following, stage S4 has 2 pipelines • Each pipeline still requires 2 cycles • Second pipeline eliminates wasted cycles • For k stages and n instructions, number of cycles = k + n

  23. Introduction to Assembly Language • MOV instruction MOV instruction copies data from one location to another. It has the following format: MOV destination, source copy source operand to destination This instruction tells the CPU to move (in reality, copy) the source operand to the destination operand. For example, MOV DX, CX copies the contents of register CX to register DX. After this instruction is executed, register DX will have the same value as register CX. The MOV instruction does not affect the source operand.

  24. MOV Examples

  25. MOV Instruction • Data can be moved among all the register as long as the source and destination registers match in size. MOV AL, DX ; Error • Data can not be moved directly into segment registers. MOV DS, 2459H ; Error • Values cannot be loaded directly into any segment register (CS, DS, ES, or SS). • To load a value into a segment register, first load it to a nonsegment register and then move it to the segment register. MOV AX, 2345H ;Ioad 2345H into AX MOV DS, AX ;load the value of AX into DS • If a value less than FFH is moved into a 16-bit register, the rest of the bits are assumed to be all zeros. MOV BX, 5 ;the result will be BX = 0005; • Moving a value that is too large into a register will cause an error. MOV BL, 7F2H ;ILLEGAL: 7F2H is larger than 8 bits MOV AX, 2FE456H ;ILLEGAL: the value is larger than AX

  26. ADD instruction • ADD instruction has the following format: ADD destination, source • The ADD instruction adds the source and destination operands and put the result in the destination. • To add two numbers such as 25H and 34H: MOV AL, 25H ;move 25 into AL MOV BL, 34H ;move 34 into BL ADD AL, BL ;AL = AL + BL • The program above can be written in many ways, MOV DH, 25H ;move 25 into DH MOV CL, 34H ;move 34 into CL ADD DH, CL ;add CL to DH; DH = DH+CL • Another way: MOV DH, 25H ;Ioad one operand into DH ADD DH, 34H ;add the second operand to DH

  27. ADD instruction • The largest number that an 8-bit register can hold is FFH. To use numbers larger than FFH (255 decimal), 16-bit registers such as AX, BX, ex, or DX must be used. • For example, to add two numbers such as 34EH and 6A5H, the following program can be used: MOV AX, 34EH ;move 34EH into AX MOV DX, 6ASH ;move 6A5H into DX ADD DX, AX ;add AX to DX • Again, any 16-bit non segment registers could have been used to perform the action above: MOV CX, 34EH ;Ioad 34EH into CX ADD CX, 6A5H ;add 6A5H to CX

  28. Flag Register in 8086/8086 • The flag register is a 16-bit register sometimes referred to as the status register • Although the register is 16 bits wide, only some of the bits are used. • The rest are either undefined or reserved by Intel. • Six of the flags are called conditional flags, meaning that they indicate some condition that resulted after an instruction was executed. • These six are CF, PF, AF, ZF, SF, and OF. • The three remaining flags are sometimes called control flags since they are used to control the operation of instructions before they are executed.

  29. Flag Register in 8086/8086

  30. EFLAGS Register in AI-32 • Status Flags • Status of arithmetic and logical operations • Control and System flags • Control the CPU operation • Programs can set and clear individual bits in the EFLAGS register

  31. Status Flags • CF, Carry Flag • Set when unsigned arithmetic result is out of range • OF, Overflow Flag • Set when signed arithmetic result is out of range • SF, Sign Flag • Copy of sign bit, set when result is negative • ZF, Zero Flag • Set when result is zero • AF, Auxiliary Carry Flag • Set when there is a carry from bit 3 to bit 4 • PF, Parity Flag • Set when parity is even • Least-significant byte in result contains even number of 1s • TF, the Trap Flag. Set to perform single-step execute • IF, Interrupt Enable Flag. indicating the external maskable interrupt requests. • DF, the Direction Flag. This bit is used to control the direction of string operations, • OF, the Overflow Flag. This flag is set whenever the result of a signed number overflow

  32. Example 1-10 • Show how the flag register is affected by the addition of 38H and 2FH. Solution MOV BH, 38H ;BH= 38H ADD BH, 2FH ;add 2F to BH, now BH=67H 38 0011 1000 + 2F 0010 1111 67 0110 0111 • CF = 0 since there is no carry beyond d7 • PF = 0 since there is an odd number of Is in the result • AF = 1 since there is a carry from d3 to d4 • ZF = 0 since the result is not zero • SF = 0 since d7 of the result is zero 1 1 1

  33. Example 1-11 • Show how the flag register is affected by MOV AL, 9CH ;AL = 9CH MOV DH, 64H ;DH = 64H ADD AL, DH ;now AL = 0 Solution 9C 1001 1100 + 64 0110 0100 1 00 1 0000 0000 • CF = 1 since there is carry beyond d7 • PF = 1 since there is an odd number of Is in the result • AF = 1 since there is a carry from d3 to d4 • ZF = 1 since the result is zero • SF = 0 since d7 of the result is zero 1 1 1 1 1

  34. The End

More Related