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  1. Overview • Problem 5.33 • Machine Language Programming Exercises • Introduction to Assembly Language Programming • Examples of Assembly Language Programming • Writing Assembly Language Programming.

  2. Homework Problem 5.33 The instructions are: x3000 AND R7, R7, #0 R7 <-0 x3001 ADD R6, R7, #1 R6 <-1 x3002 AND R4, R5, R6 Uses R6 as mask to tests one bit of R5 for a 1 x3003 BRZ x3005 Skips x3004 if bit in R5 was not a 1 x3004 ADD R0, R0, #1 Increments R0 if tested bit in R5 was a 1 x3005 ADD R6, R6, R6 Moves test bit of R6 to left x3007 ADD R7, R7, #1 Increments R7 x3008 ADD R1, R7, #-8 R1 is –7, -6, -5, -4, -3, -2, -1, 0 x3009 BRN x3002 Loops back 7 times, i.e. tests lowest 8 bits of R5 The number in R0 increases by the number of 1’s in R5. If R0 was initially 0, then R0 being 5 after completion means that there were five 1’s in the lower 8 bits of R5. If the value in R0 was initially N, then there were 5-N 1’s in the lower 8 bits of R5.

  3. Programming Exercise #1 Write a program to count the 1’s in register R0 • Flow Diagram • Machine code

  4. Programming Exercise #2 Write a program to add the contents of R0 and R1, and indicate in R2 if there was an overflow • Flow Diagram • Machine code

  5. Programming Exercise #3 Write a program to Divide the contents of R0 by the contents of R1 • Flow Diagram • Machine code

  6. Programming Exercise #4 Write a program to read characters from the keyboard, echo them on the console, and pack them into a file (2 characters per word) • Flow Diagram • Machine code

  7. LC-3 Assembly Language Syntax • Each line of a program is one of the following: • an instruction • an assember 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: LABEL OPCODE OPERANDS ;COMMENTS optional mandatory

  8. An Assembly Language Program ; ; Program to multiply a number by the constant 6 ; .ORIG x3050 LD R1, SIX LD R2, NUMBER AND R3, R3, #0 ; Clear R3. It will ; contain the product. ; The inner loop ; AGAIN ADD R3, R3, R2 ADD R1, R1, #-1 ; R1 keeps track of BRp AGAIN ; the iteration. ; HALT ; NUMBER .BLKW 1 SIX .FILL x0006 ; .END

  9. Assembler Directives • Pseudo-operations • do not refer to operations executed by program • used by assembler • look like instruction, but “opcode” starts with dot

  10. Trap Codes • LC-3 assembler provides “pseudo-instructions” foreach trap code, so you don’t have to remember them.

  11. Sample Program • Count the occurrences of a character in a file.Remember this?

  12. Count the occurrences of a character in a file (1 0f 2). ; ; 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 character file GETC ; R0 gets input character LDR R1, R3, #0 ; R1 gets first character from file ; ; 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, 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 character. LDR R1, R3, #0 ; R1 gets next char to test BRnzp TEST

  13. Count the occurrences of a character in a file (2 of 2). ; ; 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 ; ; Storage for pointer and ASCII template ; ASCII .FILL x0030 ; ASCII offset PTR .FILL x4000 ; PTR to character file .END