1 / 26

Assembly Language Review

Assembly Language Review. Being able to repeat on the Blackfin the things we were able to do on the MIPS. Assembly code things to review 50% of ENCM369 in 50 minutes. YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS Able to ADD and SUBTRACT the contents of two data registers

lderose
Download Presentation

Assembly Language Review

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. Assembly Language Review Being able to repeat on the Blackfin the things we were able to do on the MIPS Review of 50% OF ENCM369 in 50 minutes

  2. Assembly code things to review50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS • Able to ADD and SUBTRACT the contents of two data registers • Able to perform bitwise AND operations, and perform bitwise OR operations the contents of two data registers • Able to place a (small) required value or bit pattern into a data register • Able to place a (large) required value or bit pattern into a data register • Being able to write a simple “void” function (function does stuff but does not return a result) • Being able to write a simple “int” function (function does stuff and returns a result in a specified register) • Being able to ADD and SUBTRACT the contents of two memory locations • IF YOU CAN DO THE SAME THING ON THE BLACKFIN – THEN THAT’S 50% OF THE LABS AND 50% OF EXAMS MATERIAL ACED

  3. Able to ADD and SUBTRACT the contents of two data registers • It makes sense to ADD and SUBTRACT “values” stored in data registers • Blackfin DATA registers R0, R1, R2 and R3 • R0 = R1 + R2; // Additione.g. 4 + 6  10 (Decimal number) 0x14 + 0x16  0x2A (Hexadecimal number) • R3 = R1 – R2; // Subtraction e.g. 4 - 6  8 (Decimal number) 0x14 - 0x16  0xFFFFFFFE (Hexadecimal number) Review of 50% OF ENCM369 in 50 minutes

  4. Able to perform bitwise AND and OR operations on data registers • It makes sense to perform OR and AND operations on “bit-patterns” stored in data registers. • NEVER perform ADD and SUBTRACT operations on “bit-patterns” stored in data registers. • WILL THIS BE EXAMINED? THIS CAUSES A CODE DEFECT • CODE DEFECT -- your test may (accidently) get the correct answer, but your production code fails at apparently random times. • Blackfin DATA registers R0, R1, R2 and R3 • R0 = R1 & R2; // Bitwise ANDe.g. B11001100 & B01010101  B01000100 • R3 = R1 | R2; // Bitwise ORe.g. B11001100 | B01010101  B11011101 Review of 50% OF ENCM369 in 50 minutes

  5. Able to perform bitwise AND operations on data registers • KEY EMBEDDED SYSTEM OPERATION FOR CONTROLLING EMBEDDED DEVICES • For each corresponding bit in each register do • 0 & 0 = 0; 1 & 0 = 0; 0 & 1 = 0; 1 & 1 = 1 R1 = 0xCC = B 1100 1100 R2 = 0x55 = B 0101 0101 • R0 = R1 & R2; // Bitwise AND R1 = B 1 1 0 0 1 1 0 0 R2 = B 0 1 0 1 0 1 0 1 R0 = B 0 1 0 0 0 1 0 0 = 0x44 Review of 50% OF ENCM369 in 50 minutes

  6. Able to perform bitwise OR operations on data registers • KEY EMBEDDED SYSTEM OPERATION FOR CONTROLLING EMBEDDED DEVICES • For each corresponding bit in each register do • 0 | 0 = 0; 1 | 0 = 1; 0 | 1 = 0; 1 | 1 = 1 R1 = 0xCC = B 1100 1100 R2 = 0x55 = B 0101 0101 • R0 = R1 | R2; // Bitwise OR R1 = B 1 1 0 0 1 1 0 0 R2 = B 0 1 0 1 0 1 0 1 R0 = B 1 1 0 1 1 1 0 1 = 0xDD Review of 50% OF ENCM369 in 50 minutes

  7. Is it a bit pattern or a value?(Add, OR AND) Hints from “C++” If the code developer is consistent when writing the code then • Bit patterns are normally stored as “unsigned integers” e.g.unsigned int bitPattern = 0xFFA2345FF • Values are normally stored as “signed integers” e.g. signed int fooValue = -1; orint fooValue = -1; where the word “signed” is “understood”.Understood means “its there but not actually written down” (which means that it sometimes causes defects in your code – your code does not do what you expect) • Note that “bitPattern = 0xFFFFFFFF” and “fooValue = -1” are STORED as the SAME bit pattern 0xFFFFFFFFF in the registers and memory of MIPS and Blackfin processor Review of 50% OF ENCM369 in 50 minutes

  8. Being able to place a required value into a data register –Part 1 • Like the MIPS, the Blackfin uses 32 bit instructions – all registers are the same size to ensure maximum speed of the processor (highly pipelined instructions). • The 32 bit Blackfin instruction for placing a value into a data register has two parts • to have16 bits available for describing the instruction • and 16 bits for describing the “signed” 16 bit value to be put into a R0 which is “signed” 32 bit data register. • The processor has to be told whether to put the 16 bits in the top part or the bottom part of the register Review of 50% OF ENCM369 in 50 minutes

  9. Being able to place a required value into a data register –Part 1 • The 32 bit Blackfin instruction for placing a value into a data register has two parts • to have16 bits available for describing the instruction • and 16 bits for describing the “signed” 16 bit value to be put into a “signed” 32 bit data register. • This means that you have to use “2” 32-bit instructions to put large values into a data register (SAME AS MIPS). • Examples in next slides Review of 50% OF ENCM369 in 50 minutes

  10. Placing a value into a data register Similar to MIPS, different syntax • R1 = 0; legal -- 0 = 0x0000 (signed 16 bits); (becomes the signed 32 bit 0x00000000 value after auto sign extension of the 16-bit value 0x0000) • R0 = 33; legal -- 33 = 0x0021 (signed 16 bits) (becomes the signed 32 bit 0x00000021 value after auto sign extension of the 16-bit value 0x0021) • R2 = -1; legal -- -1 = 0xFFFF (signed 16 bits) (becomes the signed 32 bit 0xFFFFFFFF value after auto sign extension of the 16-bit value 0xFFFF) • R3 = -33; legal -- -33 = 0xFFDE (signed16 bits) (becomes the signed 32 bit 0xFFFFFFDE value after auto sign extension of the 16-bit value 0xFFDE) Review of 50% OF ENCM369 in 50 minutes

  11. Placing a “large” value into a data register • This approach does not work for any “large” value R1 = 40000; DOES NOT WORK WITH MIPS EITHERillegal -- as 40000 can’t be expressed as a signed 16-bit value – it is the positive 32 bit value 0x00009C40 If the assembler tried to take the bottom 16 bits of the decimal 40000 and sign extend it then this would happen “16-bit” hex value 9C40 (1001 1100 0100 0000) becomes “32-bit” hex value after sign extension 0xFFFF9C40 which is a “negative value” AND NOT WHAT YOU WANTED TO CODE • Therefore it is “illegal” to try to put a 32-bit value directly into a MIPS or a Blackfin processors (and many other processors). Review of 50% OF ENCM369 in 50 minutes

  12. Placing a “large” value into a data register • If the assembler tried to take the bottom 16 bits of the decimal 40000 and sign extend it then this would happen “16-bit” hex value 9C40 (1001 1100 0100 0000) becomes “32-bit” hex value after sign extension 0xFFFF9C40 which is a “negative value” • “illegal” just as it would be in MIPS // Want to do R1 = 40000 // Instead must do operation in two steps as with MIPS #include <blackfin.h> R1.L = lo(40000); // Tell assembler to put “bottom” // 16-bits into “low” part of R1 register R1.H = hi(40000); // Tell assembler to put “top” // 16-bits into “high” part of R1 register

  13. Placing a “large” value into a data register • A common error in the laboratory and exams is getting this two step thing “wrong” . Forgetting the second step is easy to do – just as easy to forget on Blackfin as on MIPS // Want to do R1 = 41235 – always need two MIPS or Blackfin instructions R1.L = lo(41235); // “bottom” 16-bits into “low” part of R1 register R1.H = hi(41325); // “top” 16-bits into “high” part of R1 registerTHIS SECOND STEP IS OFTEN A FORGOTTEN SECOND STEP RECOMMENDED SYNTAX TO AVOID “CODE DEFECTS”#define LARGEVALUE 41235 // C++ - like syntax R1.L = lo(LARGEVALUE); R1.H = hi(LARGEVALUE); Yes – you CAN put multiple Blackfin assembly language instructions on one line Review of 50% OF ENCM369 in 50 minutes

  14. A “void” function returns NO VALUEextern “C” void Simple_VoidASM(void) #include <blackfin.h> .section program; .global _Simple_VoidASM; _Simple_VoidASM: _Simple_VoidASM.END: RTS; // Simple example Blackfin ASM function. Stays the same in final Things in red were cut-and-pasted using the editorto save Lab. time Review of 50% OF ENCM369 in 50 minutes

  15. A simple “int” function return a valueextern “C” int Simple_IntASM(void) #include <blackfin.h> .section program; .global _Simple_IntASM; _Simple_IntASM: R0 = 7; // Return “7” _Simple_IntASM.END: RTS; Things in red were cut-and-pasted using the editor // Simple example Blackfin ASM function. Stays the same in final Review of 50% OF ENCM369 in 50 minutes

  16. Being able to ADD and SUBTRACT the contents of two memory locations Let’s set up a practical situation • A “background” code thread is putting values into an array. Processor could be MIPS or Blackfin • For “background” thread read “interrupt service routine” or ISR. • ISR work “in parallel” with the “foreground” thread that is doing the major work on the microprocessor • Write a subroutine (returns int) that adds together the first two values of this shared array Review of 50% OF ENCM369 in 50 minutes

  17. Start with a copy of the “int” function extern “C” int Simple_IntASM(void) #include <blackfin.h> .section program; .global _Simple_IntASM; _Simple_IntASM: R0 = 7; // Return “7” _SimpleInt_ASM.END: RTS; Things in red were cut-and-pasted using the editor Review of 50% OF ENCM369 in 50 minutes

  18. Modify to be extern “C” int AddArrayValuesASM(void) #include <blackfin.h> .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: R0 = 7; // Return “7” _AddArrayValuesASM.END: RTS; Things in red were cut-and-pasted using the editor Review of 50% OF ENCM369 in 50 minutes

  19. Add a “data” array in assembly code #include <blackfin.h> .section L1_data; .byte4 _fooArray[42]; // Syntax for building an array // of 32-bit values .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : R0 = 7; // Return “7” _AddArrayValuesASM .END: RTS; Things in red were cut-and-pasted using the editor Bonus marks in exams possible for ‘valid’ references to the ‘Hitch Hikers Guide’ and ‘Dr. Who’ Review of 50% OF ENCM369 in 50 minutes

  20. Plan to return “sum”, initialize sum to 0 #include <blackfin.h> .section L1_data; .byte4 _fooArray[42]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: #define sum_R0 R0 // register int sum; (Registerize int sum) sum_R0 = 0; // sum = 0; (Initialize sum) _AddArrayValuesASM .END: RTS; Things in red were cut-and-pasted using the editor Review of 50% OF ENCM369 in 50 minutes

  21. Place the memory address of the start of the array into a pointer register …. Other code .section L1_data; .byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0; #define pointer_to_array_P1 P1 // register int * pointer_to_array P1.L = lo(_fooArray); P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0]; _AddArrayValuesASM .END: RTS; Things in red were cut-and-pasted using the editor P1 is a POINTER register(address register) Review of 50% OF ENCM369 in 50 minutes

  22. Read the contents of the first array location into register R1 and add to sum_R0; …. Other code .section L1_data; .byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0; #define pointer_to_array_P1 P1 // register int * pointer_to_array P1L = lo(_fooArray); P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0];R1 = [pointer_to_array_P1]; // int temp = fooArray[0]; sum_R0 = sum_R0 + R1; // sum = sum + temp _AddArrayValuesASM.END: RTS; Things in red were cut-and-pasted using the editor Review of 50% OF ENCM369 in 50 minutes

  23. Read the contents of the second array location into register R1 and add to sum_R0; …. Other code .section L1_data; .byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0; #define pointer_to_array_P1 P1 // register int * pointer_to_array P1.L = lo(_fooArray); P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0];R1 = [pointer_to_array_P1]; // int temp = fooArray[0]; sum_R0 = sum_R0 + R1; // sum = sum + temp R1 = [pointer_to_array_P1 + 4]; // temp = fooArray[1]; sum_R0 = sum_R0 + R1; // sum = sum + temp _AddArrayValuesASM .END: RTS; Things in red were cut-and-pasted using the editor

  24. Add code to .ASM (assembly) file TO BE FIXEDCCES picturewill look similar Review of 50% OF ENCM369 in 50 minutes

  25. Assignment 1, Q1 (from 2009)Demo answer TO BE FIXEDCCES picturewill look similar Review of 50% OF ENCM369 in 50 minutes

  26. Assembly code things to review50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS • Able to ADD and SUBTRACT the contents of two data registers • Able to perform bitwise AND operations, and perform bitwise OR operations the contents of two data registers • Able to place a (small) required value or bit pattern into a data register • Able to place a (large) required value or bit pattern into a data register • Being able to write a simple “void” function (function does stuff but does not return a result) • Being able to write a simple “int” function (function does stuff and returns a result) in a specified resgister) • Being able to ADD and SUBTRACT the contents of two memory locations • IF YOU CAN DO THE SAME THING ON THE BLACKFIN – THEN THAT’S 50% OF THE LABS AND 50% OF EXAMS MATERIAL ACED

More Related