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The LC-3 – Chapter 7

The LC-3 – Chapter 7. COMP 2620. Assembly Program. ; ; 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 ;

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The LC-3 – Chapter 7

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  1. The LC-3 – Chapter 7 COMP 2620 Dr. James Money COMP 2620

  2. Assembly Program ; ; 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 GETC ; 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

  3. Assembly Program ; ; Test character for match. If a match, increment count. ; NOT R1, R1 ADD R1, R1, R0 ; If match, R1 = xFFFF NOT R1, R1 ; If match, R1 = x0000 BRnp GETCHAR ; If no match, do not increment ADD R2, R2, #1 ; ; Get next character from file. ; GETCHAR ADD R3, R3, #1 ; Point to next char 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 OUT ; ASCII code in R0 is displayed. HALT ; Halt machine

  4. Assembly Program ; ; Storage for pointer and ASCII template ; ASCII .FILL x0030 PTR .FILL x4000 .END

  5. Assembly Process • Before you can execute an LC-3 assembly program, it has to converted to machine code for the LC-3 ISA • The job of the assembler is to perform this task • Essentially, it is a translation from one assembly command to a corresponding LC-3 instruction

  6. Two-Pass Process • We consider how the assembler performs this translation to machine code • Recall there is a one-to-one correspondence between assembly instructions and machine instructions • One could do this in one pass • The first 9 lines are comments are ignored

  7. Two-Pass Process • At line 0A, we have a pseudo-op which sets the initial PC to 0x3000 • Line 0B is an AND instruction which readily translates to 0x3000: 0101010010100000

  8. Two-Pass Process • However, at line 0C, there is no knowledge yet of the memory location that PTR refers to • At this point, the assembler fails and exits if it only uses one pass of the assembly file input • To get around this, we have to use two passes of the assembly file

  9. Two-Pass Process • Pass 1: • Create list of addresses corresponding to labels • This is called a symbol table • Pass 2: • We translate the assembly instructions • We use the symbol table to clarify references

  10. Two-Pass Process • Now, at line 0C, we have to translate LD R3,PTR • But we know from the first pass, PTR refers to memory location 0x3013 • Thus, this translates to the instruction 0x3001: 0010011000010001

  11. First Pass • From our perspective, the symbol table is only list of symbolic names with 16 bit memory locations • We obtain this table by going through the file completely and see what lines and appropriate memory locations applies to each symbol

  12. First Pass • If we have made all our labels in the assembly program, then we have no unfound symbols in our table for the second pass • For now, we only consider single file programs with only one .ORIG and .END pseudo-ops

  13. First Pass • The pseudo-op at line 0A makes the current location 0x3000 • This is called the location counter or the LC • The LC is initialized to whatever value is provided by the .ORIG operand

  14. First Pass • Then, for the rest of the file, the LC increments one value for each line that is not a comment • If there is a label, an entry is put in the symbol table • The first pass ends when .END is encountered

  15. First Pass • The first label is at line 13, which is TEST • This is the fifth instruction, so LC is 0x3004 • Thus the table looks like:

  16. First Pass • The second label is at line 20 • The LC is now increments to x300B • Thus, the table is updated to

  17. First Pass • After you reach .END, the symbol table is

  18. Second Pass • In the second pass, we generate the machine code for each instruction • This time, though, we use the symbol table to assist in memory references • When we reach line 0C, we can use the address 0x3013 which corresponds to the label PTR

  19. Second Pass • The instruction is LD R3, PTR • So, the opcode is 0010 • The DR is 011 • How do we compute the PC offset?

  20. Second Pass • Recall the incremented PC is LC+1 here • Thus, PC = x3002 • And PTR is x3013 • Subtracting we have 0x3013-0x3002 = 0x11 • Thus, PC offset = 000010001 • Hence, our instruction is 0010011000010001

  21. Second Pass • Note, if the address of PTR more than +256 or less that -255, an error is issued by the assembler • This is because we only have 9 bits to encode the offset • If this is the case, you will have to use another load instruction

  22. Second Pass • Second pass continues, and finishes, incrementing LC as it goes • It uses the symbol table as needed • Resulting code is on the page 188

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