computer architecture Lecture 8 for computer science

Computer Architecture 2023-24 (WBCS010-05) Lecture 8: Assembly Reza Hassanpour r.zare.hassanpour@rug.nl

Human-Friendly Programming › Computers need binary instruction encodings... › 0001110010000110 › Humans prefer symbolic languages... › a = b + c › High-level languages allow us to write programs in clear, precise language that is more like English or math. Requires a program (compiler) to translate from symbolic language to machine instructions. › Examples: C, Python, Fortran, Java, ... 2

Assembly Language: Human-Friendly ISA Programming › Assembly Language is a low-level symbolic language, just a short step above machine instructions • Don't have to remember opcodes (ADD = 0001, NOT = 1001, ...) • Give symbolic names to memory locations -- don't have to do binary arithmetic to calculate offsets • Like machine instructions, allows programmer explicit, instruction-level specification of program › Disadvantage: › Not portable. Every ISA has its own assembly language. Program written for one platform does not run on another. 3

Assembly Language › Very similar format to instructions -- replace bit fields with symbols › For the most part, one line of assembly language = one instruction › Some additional features for allocating memory, initializing memory locations, service calls › Numerical values specified in hexadecimal (x30AB) or decimal (#10) 4 x10 is not the same as #10 !

Assembly Language Syntax › Each line of a program is either one of the following: • An instruction • An assembler directive (or pseudo-op) • A comment › Whitespace (between symbols) and case are ignored. › Comments (beginning with “;”) are also ignored. › An instruction has the following format: 6

Mandatory: Opcode and Operands › Opcodes Reserved symbols that correspond to LC-3 instructions. Listed in Appendix A and Figure 5.3. • For example: ADD, AND, LD, LDR, … › reserved means that it cannot be used as a label › Operands • Registers -- specified by Rn, where n is the register number. • Numbers -- indicated by # (decimal) or x (hex). • Label -- symbolic name of memory location (1 to 20 alphanumeric characters) • Separated by comma (whitespace ignored). • Number, order, and type correspond to instruction format. › ADD R1,R1,R3 ; DR, SR1, SR2 ADD R1,R1,#3 ; DR, SR1, Imm5 LD R6,NUMBER ; DR, address (converted to PCoffset) BRz LOOP ; nzp becomes part of opcode, address 7

Optional: Label and Comment › Label • Placed at the beginning of the line • Assigns a symbolic name to the address corresponding to that line › LOOP ADD R1,R1,#−1 ; LOOP is address of ADD BRp LOOP › Comment › A semicolon, and anything after it on the same line, is a comment › Ignored by assembler › Used by humans to document/understand programs › Tips for useful comments: • Avoid restating the obvious, as “decrement R1” • Provide additional insight, as in “accumulate product in R6” • Use comments and empty lines to separate pieces of program 8

Assembler Directive › Pseudo-operation • Does not refer to an actual instruction to be executed • Tells the assembler to do something • Looks like an instruction, except "opcode" starts with a dot Opcode Operand Meaning .ORIG address starting address of program .END end of program .BLKW n allocate n words of storage .FILL n allocate one word, initialize with value n .STRINGZ n-character string allocate n+1 locations, initialize with characters and null terminator 9

.ORIG › .ORIG tells the assembler where in memory to place the LC-3 program. › Example: .ORIG x3050 says, place the first LC-3 ISA instruction in location x3050. › If the program consists of x100 LC-3 instructions, and .ORIG says to put the first instruction in x3050, the remaining xFF instructions are placed in locations x3051 to x314F.

.END › .END tells the assembler it has reached the end of the program and need not even look at anything after it. › Any characters that come after .END will not be processed by the assembler. › .END does not stop the program during execution. › In fact, .END does not even exist at the time of execution.

.FILL › .FILL tells the assembler to set aside the next location in the program and initialize it with the value of the operand. › The value can be either a number or a label. › Example • TEN .FILL #10 › Example • .ORIG x3000 • AND R1, #0 • LOOP ADD R1, R1, #1 • ……. • FIRST .FILL LOOP

.BLKW › .BLKW tells the assembler to set aside some number of sequential memory locations (i.e., a BLocK of Words) in the program. › The actual number is the operand of the .BLKW pseudo-op. › Example › MyArray .BLKW #5

.STRINGZ › .STRINGZ tells the assembler to initialize a sequence of n+1 memory locations. › The argument is a sequence of n characters inside double quotation marks. › The first n words of memory are initialized with the zero-extended ASCII codes of the corresponding characters in the string. › Example:

.STRINGZ › For example, the code fragment › .ORIG x3010 › HELLO .STRINGZ "Hello, World!" › would result in the assembler initializing locations x3010 through x301D to the following values: › x3010: x0048 › x3011: x0065 › x3012: x006C › x3013: x006C › x3014: x006F › x3015: x002C › x3016: x0020 › x3017: x0057 › x3018: x006F › x3019: x0072 › x301A: x006C › x301B: x0064 › x301C: x0021 › x301D: x0000

Sample Program: Counting Occurrences in a File › Once again, we show the program that counts the number of times (up to nine) a user-specified character appears in a file. Count = 0 (R2 = 0) Ptr = 1st file character (R3 = M[x3012]) Input char from keybd (TRAP x23) Done? (R1 ?= EOT) Load char from file (R1 = M[R3]) Match? (R1 ?= R0) Incr Count (R2 = R2 + 1) Load next char from file (R3 = R3 + 1, R1 = M[R3]) Convert count to ASCII character (R0 = x30, R0 = R2 + R0) Print count (TRAP x21) HALT (TRAP x25) NO NO YES YES 16

Assembly Language Program 1 › ; › ; Program to count occurrences of a character in a file. › ; Character to be input from the keyboard. › ; Result to be displayed on the monitor. › ; Program only works if no more than 9 occurrences are found. › ; › ; › ; Initialization › ; › .ORIG x3000 › AND R2, R2, #0 ; R2 is counter, initially 0 › LD R3, PTR ; R3 is pointer to characters › TRAP x23 ; R0 gets character input › LDR R1, R3, #0 ; R1 gets first character › ; › ; Test character for end of file › ; › TEST ADD R4, R1, #-4 ; Test for EOT (ASCII x04) › BRz OUTPUT ; If done, prepare the output › ; › ; Test character for match. If a match, increment count. › ; › NOT R1, R1 › ADD R1, R1, #1 › ADD R1, R1, R0 ; Compute R0-R1 to compare › BRnp GETCHAR ; If no match, do not increment count › ADD R2, R2, #1 17

Assembly Language Program 2 › ; › ; Get next character from file. › ; › GETCHAR ADD R3, R3, #1 ; Point to next character. › LDR R1, R3, #0 ; R1 gets next char to test › BRnzp TEST › ; › ; Output the count. › ; › OUTPUT LD R0, ASCII ; Load the ASCII template › ADD R0, R0, R2 ; Covert binary count to ASCII › TRAP x21 ; ASCII code in R0 is displayed. › TRAP x25 ; Halt machine › ; › ; Storage for pointer and ASCII template › ; › ASCII .FILL x0030 › PTR .FILL x4000 › .END › What if we don’t put HALT (TRAP x25) at the end of the program? 18

Data or Instruction? › OUTPUT LD R0, ASCII ; Load the ASCII template › ADD R0, R0, R2 ; Covert binary count to ASCII › TRAP x21 ; ASCII code in R0 is displayed. › ; › ; Storage for pointer and ASCII template › ; › ASCII .FILL x0030 › PTR .FILL x4000 › .END › Next memory location after TRAP x21 contains x0030 › In binary: 0000 000 000110000 › Branch to PC + 48 if ? › x4000 = 0100 000 000 000000 (Jump to subroutine)

Assembly Language Program 3 › .ORIG x3000 › AND R5, R5, #0 › AND R3, R3, #0 › ADD R3, R3, #8 › LDI R1, A › ADD R2, R1, #0 › AG ADD R2, R2, R2 › ADD R3, R3, #-1 › BRnp AG › LD R4, B › AND R1, R1, R4 › NOT R1, R1 › ADD R1, R1, #1 › ADD R2, R2, R1 › BRnp NO › ADD R5, R5, #1 › NO HALT › B .FILL xFF00 › A .FILL x4000 › .END

Assembly Language Program 4 (I) › .ORIG x3000 › ONE LD R0, A › ADD R1, R1, R0 › TWO LD R0, B › ADD R1, R1, R0 › THREE LD R0, C › ADD R1, R1, R0 › ST R1, SUM › TRAP x25 › A .FILL x0001 › B .FILL x0002 › C .FILL x0003 › SUM .FILL x0004 › .END

Assembly Language Program 4 (II) › .ORIG x3000 › AND R1, R1, #0 › ONE LD R0, A › ADD R1, R1, R0 › TWO LD R0, B › ADD R1, R1, R0 › THREE LD R0, C › ADD R1, R1, R0 › ST R1, SUM › TRAP x25 › A .FILL x0001 › B .FILL x0002 › C .FILL x0003 › SUM .FILL x0004 › .END › .ORIG x3000 › AND R1, R1, #0 › ONE LD R0, A › ADD R1, R1, R0 › TWO LD R0, B › ADD R1, R1, R0 › THREE LD R0, C › ADD R1, R1, R0 › LD R0, ONE › LDI R0, ONE › ST R1, SUM › TRAP x25 › A .FILL x0001 › B .FILL x0002 › C .FILL x0003 › SUM .FILL x0004 › .END

Assembly Language Program 4 (III) › LD R0, ONE will load the value of the location shown by label ONE into R0. In this example, ONE is x3001. › Content of x3001 is LD R0, A = 0010 000 000001001 › R0 will contain x2009 › LDI R0, ONE will load the content of the location shown by an address stored at x3001. › This is equivalent to R0  M[x2009]

Assembly Process › The assembler is a program that translate an assembly language (.asm) file to a binary object (.obj) file that can be loaded into memory. › First Pass: • Scan program file, check for syntax errors • Find all labels and calculate the corresponding addresses: the symbol table › Second Pass: • Convert instructions to machine language, using information from symbol table 24

First Pass: Construct the Symbol Table 1. Find the .ORIG statement, which tells us the address of the first instruction • Initialize location counter (LC), which keeps track of the current instruction 2. For each non-empty line in the program: • If line contains a label, add label and LC to symbol table • Increment LC • NOTE: If statement is .BLKW or .STRINGZ, increment LC by the number of words allocated 3. Stop when .END statement is reached › NOTE: A line that contains only a comment is considered an empty line 25

First Pass on Sample Program (Comments Removed) › -- .ORIG x3000 › x3000 AND R2, R2, #0 › x3001 LD R3, PTR › x3002 TRAP x23 › x3003 LDR R1, R3, #0 › x3004 TEST ADD R4, R1, #-4 › x3005 BRz OUTPUT › x3006 NOT R1, R1 › x3007 ADD R1, R1, #1 › x3008 ADD R1, R1, R0 › x3009 BRnp GETCHAR › x300A ADD R2, R2, #1 › x300B GETCHAR ADD R3, R3, #1 › x300C LDR R1, R3, #0 › x300D BRnzp TEST › x300E OUTPUT LD R0, ASCII › x300F ADD R0, R0, R2 › x3010 TRAP x21 › x3011 TRAP x25 › x3012 ASCII .FILL x0030 › x3013 PTR .FILL x4000 › -- .END Label Address TEST x3004 GETCHAR x300B OUTPUT x300E ASCII x3012 PTR x3013 26

Second Pass: Convert to Machine Instructions 1. Find the .ORIG statement, which tells us the address of the first instruction. • Initialize location counter (LC), which keeps track of the current instruction 2. For each non-empty line in the program: • If line contains an instruction, translate opcode and operands to binary machine instruction. For label, lookup address in symbol table, subtract (LC+1) and replace label with that. Increment LC • If line contains .FILL, convert value/label to binary. Increment LC • If line contains .BLKW, create n copies of x0000 (or any arbitrary value). Increment LC by n • If line contains .STRINGZ, convert each ASCII character to 16-bit binary value. Add null (x0000). Increment LC by n+1 3. Stop when .END statement is reached 27

Example › .ORIG x3000 › AND R2,R2,#0 ; R2 is counter, initialize to 0 › LD R3,PTR ; R3 is pointer to characters › Set LC to x3000 › AND R2,R2,#0 ➔ 0101010010100000 › Increment LC ➔ LC = x3001 › LD R3,PTR ➔ 0010011000010001 › PTR is x3013 from Symbol table › Subtract LC+1 from x3013 ➔ x3013 – x3002 ➔ x0011 › X0011 ➔ 000010001 (9 bits binary) › Increment LC ➔ LC = x3002

Errors during Code Translation › While assembly language is being translated to machine instructions, several types of errors may be discovered • Immediate value too large -- can't fit in Imm5 field • Address out of range -- greater than LC+1+255 or less than LC+1-256 • Symbol not defined, not found in symbol table › If error is detected, assembly process is stopped and an error message is printed for the user 29

Beyond a Single Object File › Larger programs may be written by multiple programmers, or may use modules written by a third party. Each module is assembled independently, each creating its own object file and symbol table. › To execute, a program must have all of its modules combined into a single executable image › Linking is the process to combine all of the necessary object files into a single executable 30

External Symbols › In the assembly code we're writing, we may want to symbolically refer to information defined in a different module › For example, suppose we don't know the starting address of the file in our counting program. The starting address and the file data could be defined in a different module. › We want to do this: › PTR .FILL STARTofFILE › To tell the assembler that STARTofFILE will be defined in a different module, we could do something like this: › .EXTERNAL STARTofFILE › This tells the assembler that it's not an error that STARTofFILE is not defined. It will be up to the linker to find the symbol in a different module and fill in the information when creating the executable. 31

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