1. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 1 High Level Language vs. Assembly Language Data Types  HLL statements are implemented as sequences of assembly language statements  compiler converts HLL to assembly language  data types  evolved into objects a set of values –sets for different types can intersect e.g. signed & unsigned integers a set of operations

2. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 2 Data Types (Continued)  built-in data types: implemented by the language simple types: –HLL: integer, unsigned int, float, char, string, boolean, file (?) –Assembly language: byte, word, int, unsigned int –some processors also support: float – e.g. 80486+ Binary Coded Decimal (BCD) 2 digits per byte 4 bits / digit each digit range 0.. 9

3. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 3 Data Types (Continued) composite data types: –allow the design of static structures –HLL: array, record –assembly language: none (?) can reserve a block of memory use of memory is user-defined addressing modes to support access to array & record structures –user defines data structure determines set of possible values –user-defined operations

4. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 4 Data Types (Continued) HLL: other data types –enumerated –user-defined types: user defines data structure »determines set of possible values user-defined operations »how to manipulate structures –type casts

5. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 5 Data Variables, Expressions data variables  instances of objects –STATE VARIABLES! HLL: wide variety Assembly: limited (e.g. registers, labels) expressions –evaluate to a “value” –value can be either a reference to a variable a value for some type HLL: complex expressions common Assembly: [BX + DI + 3] (very limited)

6. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 6 Assignment  HLL: variable = value  left side must evaluate to a reference to a variable of some type  right side must evaluate to a legal value for a variable of the same type as lvalue  Assembly: MOV left side (lvalue) right side (rvalue)

7. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 7 Control Flow  HLL: IF statements, DO/FOR loops, GO TO, …  Assembly: JMP, conditional jumps

8. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 8 Software Interrupts  target subroutine of CALL is static must know subroutine start address when CALL is assembled  dynamic binding to subroutines? different addressing modes for CALL? –register? CALL BX – address in BX memory? –CALL [BP + 6] address received as argument?

9. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 9 Software Interrupts  alternate addressing might work in some cases what about case where software component is developed by very different group? –e.g. operating system (O/S) O/S is permanent resident of memory application is temporary resident applications need to use O/S services how can application access O/S ?

10. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 10 Software Interrupts: Interface Library  Some possible solutions (application access to O/S) 1. O/S provides interface library (.OBJ file) interface library has knowledge of addresses where O/S resides link interface library to the application library is included as part of application O/S Application interface library

11. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 11 Software Interrupts: Interface Library (Continued) 1. Interface Library Problems:  what if O/S changes?  recreate application with new library? 

12. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 12 Software Interrupts: O/S Gives I/F Address 2. O/S gives interface address when application is loaded  loader is part of O/S – knows about current resident O/S after loading application, but before passing control to application – could load BX (say) with address of O/S interface subroutine –application then : CALL BX –OR, application could save address in global variable, and then: CALL [O/S address variable]

13. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 13 Software Interrupts: O/S Gives I/F Address (Continued) 2. O/S gives interface address when application is loaded Problems: might work – but extra programming burden on application developers more complex loading process more possible places to go wrong 

14. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 14 Software Interrupts: Common Solution 3. More common solution:  like above, but existence of global variables at particular addresses is ASSUMED by developers of both O/S and applications  O/S puts pointers to O/S services in variables when O/S is loaded  application programs are developed under the assumption that these global variables exist

15. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 15 Software Interrupts: Common Solution (Cont.) shared global variable applicationO/S initialized when O/S loaded

16. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 16 Software Interrupt Instruction Software Interrupt (or TRAP) Instruction  uses the stored pointer (global variable) approach to support dynamic subroutine invocation  in p86: INT instruction  refer to subroutine invoked by INT as Interrupt Service Routine (ISR) INT Instruction  based on globally assumed vector table table of pointers to ISRs  INT has one operand: INT i operand = index into table to ISR pointer encoded as byte: range 0.. 255

17. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 17 INT Example  INT5 executes ISR whose address is stored in element 5 of the vector table –vector table starts at element 0 Do real O/S’s use the vector table approach?  YES! BIOS calls:INT 10H DOS calls: –use several table entries –e.g. INT 21H

18. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 18 Multi-Service ISR Can one ISR handle many services?  YES! have service code as one parameter ISR return-value? number and type of other parameters? –depends on the service code parameter! Example: O/S ISR pseudo-prototype: void? ISRx ( byte ServiceCode,... ??? );

19. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 19 p86 Vector Table  starts at address 0000H  entries are 4 bytes each  32-bit address low 16-bits = ISR address high 16-bits = 0000H (more in 94.203!) words stored little endian! (bytes too!)  offset to successive elements = 4 bytes  if index into vector table = i then offset to relevant entry = 0000H + 4 * i

20. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 20 p86 Vector Table: Example  INT5 accesses ISR address stored at: 0000H + 4 * 5 = 0014H 0014H ISR 5 address 0016H 0000H 0018H ISR 6 address 001AH 0000H 0010H ISR 4 address 0012H 0000H 5 th element of table 32-bit entries: stored little endian addresses: 0010H, 0011H 0012H, 0013H

21. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 21 Execution of INT Instruction  performs the following steps: 1. push FLAGS 2. push 32-bit return address –pushes high word = 0000H (94.203!) 3. IP := ISR address –gets target ISR address from mem[ 4 * i ]

22. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 22 ISR Stack Frame  the ISR stack frame is different from subroutines! parameter access? –extra bytes on stack –FLAGS –extra word of return address old BP 32-bit return address flags arguments BP return address + 8

23. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 23 IRET Instruction / Installing the ISR  Returning from an ISR: RET instruction will not work! (why not?) IRET (Interrupt Return) instruction 1. pops 32-bit return address IP := low word throw away high word (203!!) 2. pops FLAGS  INT mechanism only works if ISR address has been loaded previously into correct vector! called “installing” the ISR in the vector table

24. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 24 Summary of ISR Programming Issues  install ISR in vector table done at program initialization memory reserved for vector table ?  invoke ISR using INT extra bytes on stack! parameter access?  return from ISR using IRET restores FLAGS !

25. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 25 Example: Subroutine vs. ISR ; very beginning of program: VectorTable: ; locations 0000H-03FFH set aside for VectorTable ORG 0400H... ; near the beginning of main program ; move the address of SUBR into memory address 4*12H (= 0048H) MOV AL, 12H MOV CL, 4 MUL CL ; 8 bit operand, thus AX:= AL*CL MOV BX, AX MOV word ptr[BX], SUBR ; (cannot use AX for indirect ; addressing, and must specify word ptr to avoid ambiguity)  assume we are going to implement routine SUBR as ISR 12H  Initialization (ISR Only):

26. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 26 Example: Subroutine vs. ISR (Continued) Subroutine: CALLsubr ADDSP, ; m = 2 * # arguments ISR: INT 12H ADDSP, ; m = 2 * # arguments  Call set up

27. 7-Nov-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Oct. 2001 94201.lecture23-24-hll-interrupts 27 Example: Subroutine vs. ISR (Continued) Subroutine: SUBR: <access parameters [ BP + 4 ], [ BP + 6 ], … > RET ISR: SUBR: <access parameters [ BP + 8 ], [ BP + 10 ], … > IRET  Implementation of Routine: