High-level language structures

1. High-level language structures CAP221

2. Branching structuresIF-THEN • IF condition is true THEN execute true-branch statements End_if CAP221

3. Branching structuresIF-THEN false true condition True branch statements CAP221

4. example • Replace the number in AX by its absolute value. • Pseudo code algorithm is IF AX <0 Then replace AX by –AX END_IF CAP221

5. example • Code: CMP AX,0 ;AX<0? JNL END_IF ;no, exit , JGE NEG AX ;yes, change sign END_IF: CAP221

6. IF_THEN_ELSE • IF condition is true THEN execute true-branch statements ELSE execute FALSE-branch statements End_IF CAP221

7. IF_THEN_ELSE FALSE TRUE CONDITION FALSE BRANCH STATEMENTS TRUE BRANCH STATEMENTS CAP221

8. EXAMPLE • Suppose AL and BL contains extended ASCII characters. Display the one that comes first in the character sequence. CAP221

9. example • Pseudo code algorithm is IF AL < BL THEN display the char in AL ELSE display the char in BL END_IF CAP221

10. Code: MOV AH,2 CMP AL,BL JA ELSE_ MOV DL,AL JMP DISPLAY ELSE_: MOV DL,BL DISPLAY: INT 21H EXAMPLE CAP221

11. We use the unsigned jump because we are comparing extended characters. • JMP (unconditional jump) is needed to skip the false branch. CAP221

12. CASE • A case is a multiway branch structure that tests a register, variable, or expression for particular values or range of values. The general form is as follows: case expression values_1: statements_1 values_2: statements_2 ………….. values_n: statements_n END_CASE CAP221

13. CASE EXPRESSION VALUE_1 VALUE_N VALUE_2 ……. STATEMENTS_1 STATEMENTS_2 STATEMENTS_N CAP221

14. EXAMPLE • IF AX contains a negative number, put -1 in BX; if AX contains 0, put 0 in BX; if AX contains a positive number, put 1 in BX. CAP221

15. example • Pseudo code algorithm is Case AX < 0:put -1 in BX = 0:put 0 in BX > 0:put 1 in BX END_CASE CAP221

16. Code: CMP AX,0 JL NEGATIVE JE ZERO JG POSITIVE NEGATIVE: MOV BX,-1 JMP END_CASE ZERO: MOV BX,0 JMP END_CASE POSITIVE: MOV BX,1 END_CASE Example • NOTE: only one CMP is needed, because jump instructions do not affect the flags. CAP221

17. example • If AL contains 1 or 3, display “o”; if AL contains 2 or 4, display “e”. • Pseudo code algorithm is • Case AL 1, 3: display “o” 2, 4: display “e” End_case CAP221

18. Code: CMP AL,1 JE ODD CMP AL,3 JE ODD CMP AL,2 JE EVEN CMP AL,4 JE EVEN JMP END_ ODD: MOV DL, ’o’ JMP DIS_ EVEN: MOV DL, ’e’ DIS_: MOV AH,2 INT 21H END_: example CAP221

19. Branches with compound conditions • Sometimes the branching condition in an IF or CASE takes the form • Condition_1 and condition_2 • Condition_1 or condition_2 CAP221

20. AND conditions • An AND condition is true if and only if condition_1 and condition_2 are both true. • Likewise, if either condition is false, then the whole thing is false. CAP221

21. Example • Read a character, and if it’s an uppercase letter, display it. • Pseudo code algorithm is read a character (into AL) If (‘A’ <= character) and (character <=‘Z’) Then Display character End_if CAP221

22. Code: MOV ah,1 INT 21h CMP al,’A’ JL end_if CMP al,’Z’ JG end_if MOV DL,AL MOV AH,2 INT 21H End_if: example CAP221

23. OR conditions • condition_1 OR condition_2 is true if at least one of the conditions is true. • if both conditions are false, then the whole thing is false. CAP221

24. example • Read a character. If it’s “y” or “Y”, display it; otherwise, terminate the program. • Pseudo code algorithm is Read a character (into AL) if (character = “y” or character= “Y”) Then Display character Else Terminate the program End_if CAP221

25. MOV AH,1 INT 21h CMP AL, ’y’ JE THEN CMP AL,’Y’ JE THEN JMP ELSE_ THEN: MOV AH,2 MOV DL,AL INT 21H JMP END_IF ELSE_: MOV AH,4CH INT 21H END_IF: example CAP221

26. Looping structures • A loop is a sequence of instructions that are repeated. • For loop: this is a loop structure in which the loop statements are repeated a known number of times (a count-controlled loop). CAP221

27. For loop_count times do statementsend_for Initialize count statements Count = count - 1 Count=0 False True CAP221

28. LOOP instruction • The LOOP instruction can be used to implement a FOR loop. • General form: LOOP des_label The counter for the loop is the register CX. CAP221

29. LOOP instruction • CX is initialized to loop_count. • Execution of the LOOP instruction causes CX to be decremented automatically. • If CX is not 0, control transfers to des_label. • If CX =0, the next instruction after LOOP is done. • Des_label must precede the LOOP instruction by no more than 126 bytes. CAP221

30. LOOP instruction • So, a FOR loop can be implemented as follows. ; initialize CX to loop_count Des_label: ; body of the loop loop des_label CAP221

31. example • Write a count-controlled loop to display a row of 80 stars. • Pseudo code algorithm is For 80 times do display ‘*’ end_for CAP221

32. MOV CX,80 MOV AH,2 MOV DL,’*’ TOP: INT 21H LOOP TOP example CAP221

33. LOOP instruction • A loop instruction is executed at least once. • If CX = 0 when the loop is entered, the loop instruction causes CX to be decremented to FFFFH. • So, the loop is executed 65535 more times. CAP221

34. LOOP instruction • Solution: use the instruction JCXZ (jump if CX is zero) before the loop. JCXZ des_label CAP221

35. LOOP instruction • A loop implemented as follows is bypassed if CX is 0. JCXZ skip top: ;body of the loop loop top Skip: CAP221

36. While Loop • This loop depends on a condition. • Pseudo code while condition do statements end_while CAP221

37. While Loop condition True statements False CAP221

38. example • Write some code to count the number of characters in an input line. • Pseudo code initialize count to 0. read a character while character <> carriage_return do count=count + 1 read a character end_while CAP221

39. example MOV DX,0 ; or bx, cx, ax?? MOV AH,1 INT 21H WHILE_: CMP AL,0DH JE END_ INC DX INT 21H JMP WHILE_ END_ : CAP221

40. Repeat loop • Another conditional loop is the repeat loop. Pseudo code repeat statements until condition CAP221

41. Repeat loop statements condition False True CAP221

42. example • Write some code to read characters until a blank is read. Pseudo code Repeat read a character until character is a blank CAP221

43. example MOV AH,1 REPEAT: INT 21H CMP AL,’ ’ JNE REPEAT CAP221

44. WHILE versus REPEAT • In a while loop, the loop can be bypassed if the terminating condition is initially false. • The statements in a repeat must be done at least once. • Code for a repeat loop is likely to be shorter because there is only a conditional jump at the end, but a while loop has two jumps: a conditional jump at the top and a jmp at the bottom. CAP221

45. Programming with high-level structures • Top-down program design: The original problem is solved by solving a series of sub problems, each of which is easier to solve than the original problem. CAP221

46. example Prompt the user to enter a line of text. On the next line, display the capital letter entered that comes first alphabetically and the one that comes last. If no capital letters are entered, display “No capital letter”. sample execution: type a line of text: THE QUICK BROWN FOX JUMPED. first capital = B last capital = X CAP221

47. Top down program design First refinement: • Display the opening message. • Read and process a line of text. • Display the results. CAP221

48. example TITLE PGM6_2 : FIRST LAST CAPITALS .MODEL SMALL .STACK 100H .DATA PROMPT DB 'Type a line of text ',0dH,0AH,'$' NOCAP_MSG DB 0DH,0AH,' No Capitals $' CAP_MSG DB 0DH,0AH,'FIRST CAPITAL = ' FIRST DB '[' DB ' LAST CAPITAL = ' LAST DB '@ $' CAP221

49. Step 1 .CODE MAIN PROC ; initialize DS MOV AX,@DATA MOV DS, AX ; first step display opening message MOV AH,9 LEA DX,PROMPT INT 21H CAP221

50. Step 2: read and process a line of text Read a character. While character is not a carriage return do If character is a capital letter Then If character precedes first capital Then First capital = character End_if If character follows last capital Then Last capital = character End_if End_if Read a character End_while CAP221