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Comparing and Branching

Comparing and Branching. if s and loops Part B. loops. Recall the forever loop. for ( ; ; ) { … }. lp: … jmp lp. while loop. while (i < 100) { … }. How can we do this in Assembler?. while loop. while (i < 100) { … }. lp: cmp eax, 100 jge done ;or jnl …

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Comparing and Branching

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  1. Comparing and Branching ifs and loops Part B

  2. loops

  3. Recall the forever loop for ( ; ; ) { … } lp: … jmp lp

  4. while loop while (i < 100) { … } How can we do this in Assembler?

  5. while loop while (i < 100) { … } lp: cmp eax, 100 jge done ;or jnl … jmp lp done:

  6. while loop i = 0; while (i < 100) { … ++i; } How can we do this in Assembler?

  7. while loop i = 0; while (i < 100) { … ++i; } mov eax, 0 lp: cmp eax, 100 jge done ;or jnl … inc eax jmp lp done:

  8. for loop for (int i=0; i<100; i++) { … } How can we accomplish this in Assembler? But first, when is the condition evaluated?

  9. for loop for (int i=0; i<100; i++) { … } How can we accomplish this in Assembler? But first, when is the condition evaluated? Before we enter the body of the loop.

  10. for loop for (int i=0; i<100; i++) { … } ;same as the last while loop! mov eax, 0 lp: cmp eax, 100 jge done ;or jnl … inc eax jmp lp done: i = 0; while (i < 100) { … ++i; }

  11. for loop for (int i=10; i>0; i--) { … } How can we accomplish this in Assembler?

  12. for loop for (int i=10; i>0; i--) { … } mov eax, 10 lp: cmp eax, 0 jle done ;or jng … dec eax jmp lp done:

  13. do-while loop int i = 7; do { … i--; while (i > 2); mov eax, 7 lp: … dec eax cmp eax, 2 jg lp They are the same, for a change!

  14. LOOP instruction • Decrements ecx. • Loops (branches) if ecx is not equal to zero.

  15. LOOP instruction int i = 7; do { … i--; while (i !=0); ;MUST be ecx mov ecx, 7 lp: … loop lp

  16. LOOP instruction • The LOOP instruction is NOT a direct replacement for the for loop. for (int i=7; i>0; i--) { … } • How is it different from the for loop? ;MUST be ecx mov ecx, 7 lp: … loop lp

  17. LOOP instruction • The LOOP instruction is NOT a direct replacement for the for loop. for (inti=7; i>0; i--) { … } • How is it different from the for loop? • Test is at end. • Test is ecx != 0. ;MUST be ecx mov ecx, 7 lp: … loop lp

  18. LOOP instruction How can we do this in Assembler w/out the loop instruction? mov ecx, 7 lp: … loop lp

  19. LOOP instruction mov ecx, 7 lp: … dec ecx jnz lp How can we do this in Assembler w/out the loop instruction? mov ecx, 7 lp: … loop lp What do you think is faster?

  20. LOOP instruction mov ecx, 7 lp: … dec ecx jnz lp How can we do this in Assembler w/out the loop instruction? mov ecx, 7 lp: … loop lp Let’s see! What do you think is faster?

  21. utilities • The utilities.asm file (on the course’s software web page) contains the dump procedure and two other procedures that can be used to time instructions: • call resetTime • resets the timer to zero and starts the timer running • call reportTime • reports the elapsed time in milliseconds • Only one timer can be active at a time (no nested timers).

  22. LOOP vs. DEC/JG instruction timing • nop, nop, dec • repeat above via LOOP (2,000,000,000 times) • required 12,110 milliseconds total • ~6 ns each • repeat above via DEC/JG (2,000,000,000 times) • required 4,047 milliseconds total • ~2 ns each • Conclusion: DEC/JG is 3x faster than LOOP!

  23. Single NOP instruction timing • nop, nop, dec • repeat above via DEC/JG (2,000,000,000 times) • required 4,047 milliseconds total • ~2 ns each • nop,nop, dec • repeat above via DEC/JG (2,000,000,000 times) • required 2,890 milliseconds total • ~1.5ns each • Conclusion: Each nop requires ~0.5ns.

  24. A note regarding code line labels • It becomes onerous to continually make up unique line labels. So . . . • @@: • Defines a code label recognizable only between label1 and label2, where label1 is either start of code or the previous @@: label, and label2 is either end of code or the next @@: label. (See @B and @F.) • @B • Refers to the location of the previous @@: label (backward). • @F • Refers to the location of the next @@: label (forward).

  25. A note regarding code line labels @@: • Defines a code label recognizable only between label1 and label2, where label1 is either start of code or the previous @@: label, and label2 is either end of code or the next @@: label. (See @B and @F.) @B • Refers to the location of the previous @@: label. @F • Refers to the location of the next @@: label. mov ecx, 7 lp: … dec ecx jnz lp How can we accomplish this w/out lp?

  26. A note regarding code line labels @@: • Defines a code label recognizable only between label1 and label2, where label1 is either start of code or the previous @@: label, and label2 is either end of code or the next @@: label. See @B and @F. @B • Refers to the location of the previous @@: label. @F • Refers to the location of the next @@: label. mov ecx, 7 lp: … dec ecx jnz lp We can accomplish this w/out lp: mov ecx, 7 @@: … dec ecx jnz @b

  27. A note regarding code line labels @@: • Defines a code label recognizable only between label1 and label2, where label1 is either start of code or the previous @@: label, and label2 is either end of code or the next @@: label. See @B and @F. @B • Refers to the location of the previous @@: label. @F • Refers to the location of the next @@: label. ; for (int i=0; i<100; i++) { mov eax, 0 lp: cmp eax, 100 jge done … inc eax jmp lp done: How can we accomplish this w/out lp and done?

  28. A note regarding code line labels @@: • Defines a code label recognizable only between label1 and label2, where label1 is either start of code or the previous @@: label, and label2 is either end of code or the next @@: label. See @B and @F. @B • The location of the previous @@: label. @F • The location of the next @@: label. ; for (int i=0; i<100; i++) { mov eax, 0 lp: cmp eax, 100 jge done … inc eax jmp lp done: We can accomplish this w/out lp and done: mov eax, 0 @@: cmp eax, 100 jge @f … inc eax jmp @b @@:

  29. Quiz: Nested for loops #1 • Write the following in Assembler. for (int i=0; i<100; i++) { for (int j=1; j<=10; j++) { … } }

  30. Nested for loops #2 • Write the following in Assembler. for (int i=0; i<100; i++) { for (int j=10; j>=1; j--) { … } }

  31. Nested for loops #3 • Write the following in Assembler. for (int i=0; i<100; i++) { for (int j=1; j<=i; j+=3) { … } }

  32. Java expressions #4 • Convert the following to Assembler: int i = 100; int j = 1 + 3 * i; You must define i and j.

  33. Disjunction #5 • Convert the following to Assembler: if ( f() || g() || h() ) { //disjuncts (short circuits) a = 10; } else { b = 99; } Note(s): Assume a and b are 32-bit integers and f, g, and h are boolean functions that are already defined for you. f, g, and h return values of 1 in eax to indicate true, 0 to indicate false.

  34. Disjunction #6 • Convert the following to Assembler: if ( f() & g() & a<100 ) { //does not disjunct (short circuit) a = 10; } else { b = 99; } Note(s): Assume a and b are 32-bit integers and f, g, and h are boolean functions that are already defined for you. f, g, and h return values of 1 in eax to indicate true, 0 to indicate false.

  35. Operator precedence • The closer to the top of the table an operator appears, the higher its precedence. • Operators with higher precedence are evaluated before operators with relatively lower precedence. • Operators on the same line have equal precedence. When operators of equal precedence appear in the same expression, a rule must govern which is evaluated first. All binary operators except for the assignment operators are evaluated from left to right; assignment operators are evaluated right to left. • http://download.oracle.com/javase/tutorial/java/nutsandbolts/operators.html

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