Transcript Part Two
n bits first word second word • • • i th word • • • last word Figure 2.5. Memory words. 32 bits b1 • • • b31 b30 b0 Sign bit: b31= 0 for positive numbers b31= 1 for negative numbers (a) A signed integer 8 bits 8 bits 8 bits 8 bits ASCII character ASCII character ASCII character ASCII character (b) Four characters Figure 2.6. Examples of encoded information in a 32-bit word. Address Begin execution here i i +4 i +8 Contents Move Add Move A,R0 B,R0 R0,C 3-instruction program segment A B C Figure 2.8. A program for C [A] + [B]. Data for the program i i + 4 i + 8 i + 4n - 4 i + 4n SUM NUM1 NUM2 NUM n Move Add Add Add Move NUM1,R0 NUM2,R0 NUM3,R0 • • • NUM n ,R0 R0,SUM • • • • • • Figure 2.9. A straight-line program for adding n numbers. Move Clear LOOP Program loop N,R1 R0 Determine address of "Next" number and add "Next" number to R0 Decrement Branch>0 Move SUM N NUM1 NUM2 NUM n R1 LOOP R0,SUM • • • n • • • Figure 2.10. Using a loop to add n numbers. Add (R1),R0 Add (A),R0 Main memory B R1 Operand B Register (a) Through a general-purpose register A B B Operand (b) Through a memory location Figure 2.11. Indirect addressing. Address LOOP Contents Move Move Clear Add Add Decrement Branch>0 Move N,R1 #NUM1,R2 R0 (R2),R0 #4,R2 R1 LOOP R0,SUM Initialization Figure 2.12. Use of indirect addressing in the program of Figure 2.10. Add 20(R1),R2 1000 1000 R1 20 R1 20 = offset 1020 Operand (a) Offset is given as a constant Add 1000(R1),R2 1000 20 = offset 1020 Operand (b) Offset is in the index register Figure 2.13. Indexed addressing. n Student ID Test 1 Test 2 N LIST LIST + 4 LIST + 8 LIST + 12 LIST + 16 Test 3 Student ID Test 1 Test 2 Test 3 • • • Figure 2.14. A list of students' marks. Student 1 Student 2 LOOP Move #LIST,R0 Clear R1 Clear R2 Clear R3 Move N,R4 Add 4(R0),R1 Add 8(R0),R2 A dd 12(R0),R3 Add #16,R0 Decrement R4 Branch>0 LOOP Move R1,SUM1 Move Move R2,SUM2 R3,SUM3 Figure 2.15. Indexed addressing used in accessing test scores in the list in Figure 2.14. LOOP Move Move Clear Add Decrement Branch>0 Move N,R1 #NUM1,R2 R0 (R2)+,R0 R1 LOOP R0,SUM Initialization Figure 2.16. The Autoincrement addressing mode used in the program of Figure 2.12. LOOP 100 104 Move Move N,R1 #NUM1,R2 108 112 116 Clear Add Add R0 (R2),R0 #4,R2 120 124 128 132 Decrement Branch>0 Move R1 LOOP R0,SUM SUM N 200 204 NUM1 NUM2 208 212 NUM n 604 100 Figure 2.17. Memory arrangement for the program in Figure 2.12. Memory address label Assembler directives SUM N NUM1 Statemen ts that generate machine instructions Assemblerdirectives START LOOP Operation EQU ORIGIN DATAWORD RESERVE ORIGIN MOVE MOVE CLR ADD ADD DEC BGTZ MOVE RETURN END Addressing or data information 200 204 100 400 100 N,R1 #NUM1,R2 R0 (R2),R0 #4,R2 R1 LOOP R0,SUM START Figure 2.18. Assembly language representation for the program in Figure 2.17. Figure 2.19 Bus connection for processor , keyboard, and display . Move #LOC,R0 READ TestBit Branch=0 MoveByte #3,INSTATUS READ DATAIN,(R0) ECHO TestBit Branch=0 MoveByte #3,OUTSTATUS ECHO (R0),DATAOUT Compare #CR,(R0)+ Branch0 READ Initialize pointer registerR0 to point to the addressof the first locationin memory wherethe charactersare to be stored. Wait for a characterto be entered in the keyboard buffer DATAIN. Transferthe characterfrom DATAIN into the memory(this clears SIN to 0). Wait for the display to becomeready. Move thecharacterjust read to the display buffer register(this clearsSOUT to 0). Check if the characterjust read is CR (carriagereturn). If it is not CR, then branch back and read anothercharacter. Also, increment the pointer to store the next character. Figure 2.20. A program that reads a line of characters and displays it.