Lecture 9 examples (1)
- 1. Lecture 9 Memory Accessible addresses and their Examples Iqra university
- 2. Revision, 8086 has a 20-bit address bus & thus can address a maximum of 1MB (2^20) of memory But 8086 has 16-bit registers which can hold only 16-bit addresses Solution is to divide 1MB memory in 64KB segments (2^16)(total sixteen) as they can be accessed by using 16-bit addresses. However, only 4 segments are accessible by 8086 at a time. These 4 segments are: Code segment: Used for storing the codes or instructions Stack segment: Used as a stack • • • • Data segment: Used for storing data Extra segment: Used for storing in-excess data 2^20=1MB=1048576 2^16=65536= 64KB 1048576/65536=16
- 4. • Accessing to memory location there are four addresses that you all have to understand • Logical address • Segment address • Offset • Physical address /Effective address Memory Accessing Addresses
- 5. Logical Address • Logical address is implemented practically by compiler through different emulators or • Logical Address is generated by CPU while a program is running. • The logical address is virtual address as it does not exist physically, therefore, it is also known as Virtual Address. • Logical address = Segment Address : Offset
- 6. Contd. Last byte 2FFFFH Logical Address= 2222:0016
- 7. Segment address (Base Address of segment) • segment address is a 4-digit hexadecimal address which points to the start of a 64 kB chunk (a single segment )of data. • Segment address relates to base address of a single memory segments • As memory is of 1MB= 2^20 bytes so segment address is of 20 bits as 00000 to FFFFFH • Where, segment register are bit DS, CS, SS, ES • Segment register hold only upper 16 bits that is 0000H to FFFFH
- 8. 1 Code Segment 3 4 Data Segment Extra Segment 7 8 9 10 11 12 13 14 15 Stack Segment Memory 00000H FFFFFH 1MB Address Range StartingAddresses ofSegments 1000 0H 4000 0H 0H5000 F000 0H CSR DSR ESR SSR We can take any value of starting address of a segments but it must be divisible by 16 and lowest 4 bits (lowest hex digit) must be 0 Segment registers (CSR, DSR, SSR & ESR) hold upper 16-bits starting of 20-bits address of segments SEGMENT REGISTERS
- 9. Contd. Last byte 2FFFFH Segment Address Logical Address= 2222:0016
- 10. Offset • The offset is also a 4-digit hexadecimal address which defines the displacement from the segment base pointer • Offset address indicates displacement address that is address required to move in 64KB memory segment. • Offset is of 16 bits 0000H to FFFFH
- 11. Contd. Last byte 2FFFFH Segment Address Logical Address= 2222:0016
- 12. All offsets are limited to 16-bits. It means that the maximum size possible for segment is 2^16= 65,535 bytes (64 KB). The offset of the first location within the segment is 0000 H. The offset of the last location in segment is FFFF H Limitation of offset address
- 13. Physical Address • Physical Address identifies a physical location of required data in a memory. The user never directly deals with the physical address but can access by its corresponding logical address. • Physical Address is calculated by adding segment address (appended by 0H ) to the offset.
- 14. Physical Address (20 Bits) Adder Segment Register (16 bits) 0 0 0 0 Offset Value (16 bits) Physical Address Generation in 8086 Actual address for memory Upper 16bit of starting address of a segment 20-bits 16-bits Points to a memory location within a segment
- 15. Contd. Last byte 2FFFFH Segment Address Logical Address= 2222:0016 Physical address
- 16. Question: How is a 20-bit address obtained if there are only 16- bit registers???????? The answer lies in the next few slides. The 20-bit address of a byte is called its Physical Address. But, it is specified as a Logical Address. Logical address is in the form of: Segment address(Base Address) : Offset Offset is the displacement of the memory location from the starting location of the segment.
- 17. Important points to remember: Address of a segment is of 20-bits, however, within a segment a memory location can be specified or accessed using a 16-bit address called offset • Physical address is generated for accessing memory locations. Where Physical address will be calculated by adding segment address to offset.
- 18. If the data at any location has a logical address specified as: 2222 H : 0016 H Ques :What will be its physical address???? Example
- 19. Contd. Logical Address 2222 H : 0016 H
- 20. • The number 0016 H is the offset. • 2222 H is the value of DS. To convert this 16-bit address into 20-bit, the BIU (Bus interface unit )appends 0H to the LSBs of the address. After appending, the starting address of the Data Segment becomes 22220H. Then, it adds offset to it 22220+0016. Solution:
- 21. Therefore Physical Address: = 22220 H + 0016 H 22236 H Contd.
- 23. Where to look offset for respective segment Default Registers Assigned to store offset values for different segments
- 24. Offset is derived from the combination of pointer registers, index registers the Instruction Pointer, and immediate values (called displacement) 0000 + Segment address Offset Memory address Examples CS IP+ Instruction (code) address 5 0 0 0 0 F F E 0 5 F F E 0 SS SP + Stac k add r ess 3 4 8 A 0 4 2 1 4 3 8 A B 4 1 2 3 4 0 0 0 2 2 1 2 3 6 2 DS DI + Data addr ess Physical Address Calculation
- 25. 1 Data Segment 3 4 Code Segment Extra Segment 7 8 9 10 11 12 13 14 15 Stack Segment Memory 00000H FFFFFH 1MB Address Range 348A0H + 4214 H 38AB4 H CS IP Physical Address Start of Code Segment (348A0H) Code Byte at38AB4H IP = 4214H Example of Physical Address Generation for Code Segment
- 26. Memory Segment Register Offset Physical Address + 0050 DS: SI 05C0 0050 05C00H 05C50H 05C0 0 05C50H Data is fetched with respect to the DS register which contains starting or base address The offset is in SI (default register for DS) DS:EA 0H 0FFFFFH Example of Physical Address Generation for Data Segment
- 27. Conclusions • A 16-bit address bus addresses up to 64KB of memory from 0000h to FFFFh • a 20-bit address bus addresses a total of 1 MB from 00000h to FFFFFh. • The 80386/80486/Pentium processors have a 32-bit address bus and can address from 00000000h to FFFFFFFFh. • The 20-bit physical address is generated by: • Adding 16-bit contents of a segment register with an 16-bit offset value • Offset is stored in a corresponding default register (either in IP, BX, SI, DI, BP or SP. Different segments have different default register for offset, • BIU always appends 4 zeros automatically to the 16-bit address of a segment register (to make it 20-bit) because it knows the starting address of a segment always ends with 4 zeros
- 28. The contents of the following registers are: CS = 111 H DS = 3333 H SS = 2526 H IP = 1232 H SP = 1100 H DI = 0020 H Calculate the corresponding physical addresses for the address bytes in CS, DS and SS. Examples to be solved on VLE
- 29. • Thank you for listening • Queries will be answered on VLE