Assembly Language Review Being able to repeat on the Blackfin the things we were able to do on the MIPS 3/3/2016 Review of 50% OF ENCM369 in 50 minutes1

Assembly code things to review 50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS Being able to ADD and SUBTRACT the contents of two data registers Being able to bitwise AND and bitwise OR the contents of two data registers Being able to place a (small) required value into a data register Being able to place a (large) required value into a data register Being able to write a simple “void” function (returns nothing) Being able to write a simple “int” function (returns and int) 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 ACED 3/3/ / 28

Being 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;// Addition e.g  10 (Decimal) 0x14 + 0x16  0x2A (Hexadecimal) R3 = R1 – R2;// Subtraction e.g  8 (Decimal) 0x14 - 0x16  0xFFFFFFFE (Hexadecimal) 3/3/2016 Review of 50% OF ENCM369 in 50 minutes3 / 28 CHANGED SLIDE

Being able to bitwise AND and OR the contents of two 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. (Although SOMETIMES get the correct answer – code defect) Blackfin DATA registers R0, R1, R2 and R3 R0 = R1 & R2;// Bitwise AND e.g. B & B  B R3 = R1 | R2;// Bitwise OR e.g. B | B  B /3/2016 Review of 50% OF ENCM369 in 50 minutes4 / 28 CHANGED SLIDE

Is it a bit pattern or a value? 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; or int 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) 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 3/3/2016 Review of 50% OF ENCM369 in 50 minutes5 / 28 CHANGED SLIDE

Being able to place a required value into a data register –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 “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). 3/3/2016 Review of 50% OF ENCM369 in 50 minutes6 / 28

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 0x after auto sign extension of the 16-bit value 0x0000) R0 = 33; legal = 0x0021 (signed 16 bits) (becomes the signed 32 bit 0x after auto sign extension of the 16-bit value 0x0021) R2 = -1; legal = 0xFFFF (signed 16 bits) (becomes the signed 32 bit 0xFFFFFFFF after auto sign extension of the 16-bit value 0xFFFF) R3 = -33; legal = 0xFFDE (signed16 bits) (becomes the signed 32 bit 0xFFFFFFDE after auto sign extension of the 16-bit value 0xFFDE) 3/3/2016 Review of 50% OF ENCM369 in 50 minutes7 / 28

Placing a “large” value into a data register This approach does not work for any “large” value R1 = 40000; DOES NOT WORK WITH MIPS EITHER illegal -- as 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 and sign extend it then this would happen “16-bit” hex value 9C40 ( ) becomes “32-bit” hex value after sign extension 0xFFFF9C40 which is a “negative value” Therefore it is “illegal” to try to put a 32-bit value directly into a register; just as it would be illegal to try in MIPS. 3/3/2016 Review of 50% OF ENCM369 in 50 minutes8 / 28

Placing a “large” value into a data register If the assembler tried to take the bottom 16 bits of the decimal and sign extend it then this would happen “16-bit” hex value 9C40 ( ) 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 = // Instead must do operation in two steps as with MIPS #include 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 3/3/ / 28

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 = 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 register 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 3/3/2016 Review of 50% OF ENCM369 in 50 minutes10 / 28

A “void” function returns NO VALUE extern “C” void SimpleVoidASM(void) #include.section program;.global _SimpleVoidASM; _SimpleVoidASM: _SimpleVoidASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes11 / 28 Things in red were cut-and-pasted using the editor to save Lab. time

A simple “int” function return a value extern “C” int SimpleIntASM(void) #include.section program;.global _SimpleIntASM; _SimpleIntASM: R0 = 7; // Return “7” _SimpleIntASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes12 / 28 Things in red were cut-and-pasted using the editor

Being able to ADD and SUBTRACT the contents of two memory locations Let’s set up a practical situation A “background” 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 3/3/2016 Review of 50% OF ENCM369 in 50 minutes13 / 28

Start with a copy of the “int” function extern “C” int SimpleIntASM(void) #include.section program;.global _SimpleIntASM; _SimpleIntASM: R0 = 7; // Return “7” _SimpleIntASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes14 / 28 Things in red were cut-and-pasted using the editor

Modify to be extern “C” int AddArrayValuesASM(void) #include.section program;.global _AddArrayValuesASM; _AddArrayValuesASM: R0 = 7; // Return “7” _AddArrayValuesASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes15 / 28 Things in red were cut-and-pasted using the editor

Add a “data” array #include.section L1_data;.byte4 _fooArray[2]; // Syntax for building an array // of 32-bit values.section program;.global _AddArrayValuesASM; _AddArrayValuesASM : R0 = 7; // Return “7” _AddArrayValuesASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes16 / 28 Things in red were cut-and-pasted using the editor

Plan to return “sum”, initialize sum to 0 #include.section L1_data;.byte4 _fooArray[2];.section program;.global _AddArrayValuesASM; _AddArrayValuesASM: #define sum_R0 R0// register int sum; sum_R0 = 0; // sum = 0; _AddArrayValuesASM.END: RTS; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes17 / 28 Things in red were cut-and-pasted using the editor

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; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes18 / 28 Things in red were cut-and-pasted using the editor P1 is a POINTER register (address register)

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; 3/3/2016 Review of 50% OF ENCM369 in 50 minutes19 / 28 Things in red were cut-and-pasted using the editor

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; 3/3/ / 28 Things in red were cut-and-pasted using the editor

Add code to.ASM (assembly) file 3/3/2016 Review of 50% OF ENCM369 in 50 minutes21 / 28

Assignment 1, Q1 Demo answer 3/3/2016 Review of 50% OF ENCM369 in 50 minutes22 / 28

Assembly code things to review 50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS Being able to ADD and SUBTRACT the contents of two data registers Being able to bitwise AND and bitwise OR the contents of two data registers Being able to place a (small) required value into a data register Being able to place a (large) required value into a data register Being able to write a simple “void” function (returns nothing) Being able to write a simple “int” function (returns and int) 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 ACED 3/3/ / 28