1. Binary Operations Math/Logical

2. Binary Math

3. Decimal Addition Example 3 7 5 8 + 4 6 5 7 1) Add 8 + 7 = 15 Write down 5, carry 1 1 8 11 415 4) Add 3 + 4 + 1 = 8 Write down 8 3) Add 7 + 6 + 1 = 14 Write down 4, carry 1 2) Add 5 + 5 + 1 = 11 Write down 1, carry 1 Add 3758 to 4657:

4. Decimal Addition Explanation 1 1 1 3 7 5 8 + 4 6 5 7 8 4 1 5 What just happened? 1 1 1 (carry) 3 7 5 8 + 4 6 5 7 8 14 11 15 (sum) - 10 10 10 (subtract the base) 8 4 1 5 So when the sum of a column is equal to or greater than the base, we subtract the base from the sum, record the difference, and carry one to the next column to the left.

5. Binary Addition Rules Rules: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1(just like in decimal) 1 + 1= 2 10 = 10 2 = 0 with 1 to carry 1 + 1 + 1= 3 10 = 11 2 = 1 with 1 to carry

6. Binary Addition Example 1 1 1 0 1 1 1 + 0 1 1 1 0 0 1 1 111 010011 Example 1: Add binary 110111 to 11100 Col 1) Add 1 + 0 = 1 Write 1 Col 2) Add 1 + 0 = 1 Write 1 Col 3) Add 1 + 1 = 2 (10 in binary) Write 0, carry 1 Col 4) Add 1+ 0 + 1 = 2 Write 0, carry 1 Col 6) Add 1 + 1 + 0 = 2 Write 0, carry 1 Col 5) Add 1 + 1 + 1 = 3 (11 in binary) Write 1, carry 1 Col 7) Bring down the carried 1 Write 1

7. Binary Addition Explanation 1 1 0 1 1 1 + 0 1 1 1 0 0 -. 1 1 111 010011 In the first two columns, there were no carries. In column 3, we add 1 + 1 = 2 Since 2 is equal to the base, subtract the base from the sum and carry 1. In column 4, we also subtract the base from the sum and carry 1. In column 6, we also subtract the base from the sum and carry 1. In column 5, we also subtract the base from the sum and carry 1. In column 7, we just bring down the carried 1 2 2223 222 What is actually happened when we carried in binary?

8. Binary Addition Verification Verification 110111 2  55 10 +011100 2 + 28 10 83 10 64 32 16 8 4 2 1 1 0 1 0 0 1 1 = 64 + 16 + 2 +1 = 83 10 1 1 0 1 1 1 + 0 1 1 1 0 0 1010011 You can always check your answer by converting the figures to decimal, doing the addition, and comparing the answers.

9. Binary Addition Example 2 Verification 111010 2  58 10 +001111 2 + 15 10 73 10 64 32 16 8 4 2 1 1 0 0 1 0 0 1 = 64 + 8 +1 = 73 10 1 1 1 0 1 0 + 0 0 1 1 1 1 1 1 11 001010 Example 2: Add 1111 to 111010. 11

10. Decimal Subtraction Example 8 0 2 5 - 4 6 5 7 Subtract 4657 from 8025: 791 1 11 8633 1)Try to subtract 5 – 7  can’t. Must borrow 10 from next column. 4) Subtract 7 – 4 = 3 3) Subtract 9 – 6 = 3 2)Try to subtract 1 – 5  can’t. Must borrow 10 from next column. But next column is 0, so must go to column after next to borrow. Add the borrowed 10 to the original 0. Now you can borrow 10 from this column. Add the borrowed 10 to the original 5. Then subtract 15 – 7 = 8. Add the borrowed 10 to the original 1.. Then subract 11 – 5 = 6

11. Decimal Subtraction Explanation  So when you cannot subtract, you borrow from the column to the left.  The amount borrowed is 1 base unit, which in decimal is 10.  The 10 is added to the original column value, so you will be able to subtract. 8633 8 0 2 5 - 4 6 5 7

12. Binary Subtraction Explanation  In binary, the base unit is 2  So when you cannot subtract, you borrow from the column to the left.  The amount borrowed is 2.  The 2 is added to the original column value, so you will be able to subtract.

13. Binary Subtraction Example 1 1 1 0 0 1 1 - 1 1 1 0 0 Example 1: Subtract binary 11100 from 110011 2 002 1 2 1101 Col 1) Subtract 1 – 0 = 1 Col 5) Try to subtract 0 – 1  can’t. Must borrow from next column. Col 4) Subtract 1 – 1 = 0 Col 3) Try to subtract 0 – 1  can’t. Must borrow 2 from next column. But next column is 0, so must go to column after next to borrow. Add the borrowed 2 to the 0 on the right. Now you can borrow from this column (leaving 1 remaining). Col 2) Subtract 1 – 0 = 1 Add the borrowed 2 to the original 0. Then subtract 2 – 1 = 1 1 Add the borrowed 2 to the remaining 0. Then subtract 2 – 1 = 1 Col 6) Remaining leading 0 can be ignored.

14. Binary Subtraction Verification Verification 110011 2  51 10 - 11100 2 - 28 10 23 10 64 32 16 8 4 2 1 1 0 1 1 1 = 16 + 4 + 2 + 1 = 23 10 1 1 0 0 1 1 - 1 1 1 0 0 2 002 1 2 11011 Subtract binary 11100 from 110011:

15. Binary Subtraction Example 2 1 0 1 0 0 1 - 1 0 1 0 0 Example 2: Subtract binary 10100 from 101001 2002 11010 Verification 101001 2  41 10 - 10100 2 - 20 10 21 10 64 32 16 8 4 2 1 1 0 1 0 1 = 16 + 4 + 1 = 21 10

16. Logical, Shift, and Rotate Operations

17. Logical, Shift and Rotate Operations  A particular bit, or set of bits, within the byte can be set to 1 or 0, depending on conditions encountered during the execution of a program.  When so used, these bits are often called "flags".  Frequently, the programmer must manipulate these individual bits - an activity sometimes known as "bit twiddling".  The logical, shift, and rotate operations provide the means for manipulating the bits.

18. Logical OR Rules OR Operations OR Results in 1 if either or both of the operands are 1. OR Table 0 OR 0 = 0 0 OR 1 = 1 1 OR 0 = 1 1 OR 1 = 1

19. Logical OR Operation To perform the OR operation, take one column at a time and perform the OR operation using the OR table. Ex 1: 1 0 0 1 0 0 1 1 OR0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1

20. Logical OR Examples Ex 3: 0 1 1 1 OR0 0 1 0 0 1 1 1 Ex 2: 1 1 0 0 1 0 0 1 OR0 0 0 0 1 0 1 0 1 1 0 0 1 0 1 1

21. Logical XOR Rules XOR Operations The exclusive OR. Similar to OR except that it now gives 0 when both its operands are 1. Rules. 0 XOR 0 = 0 0 XOR 1 = 1 1 XOR 0 = 1 1 XOR 1 = 0

22. Logical XOR Examples Ex 1:1 0 0 1 1 0 0 1 XOR0 0 0 0 1 1 1 1 1 0 0 1 0 1 1 0 Ex 2: 0 1 1 1 XOR0 0 1 0 0 1 0 1

23. Logical AND Rules AND Operations AND yields 1 only if both its operands are 1. Rules. 0 AND 0 = 0 0 AND 1 = 0 1 AND 0 = 0 1 AND 1 = 1

24. Logical AND Examples Ex 1: 1 1 0 1 0 0 1 1 AND 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 Ex 2: 0 1 1 1 AND 1 0 0 1 0 0 0 1

25. Logical NOT NOT Operations NOT is a separate operator for flipping the bits. Rules. NOT 0 = 1 NOT 1 = 0 Example.NOT1 0 1 0 = 0 1 0 1

26. Shift and Rotate operations Whereas logical operations allow the changing of bit values in place, SHIFT and ROTATE operations allow bits to be moved left or right without changing their values.

27. Shift Left operation SHL SHL (shift left) shifts each bit one place to the left. The original leftmost bit is lost and a 0 is shifted into the rightmost position. Ex 1.SHL1 1 0 1 Ex 2.SHL1 1 0 0 = 1 0 0 0 01 1 0 1 = 1 0 1 0

28. Shift Left - Multiple Bits SHL # bits means to shift left # times Ex 1: SHL 31 0 0 1 1 1 0 0 Ex 2: SHL 21 1 1 0 0 1 1 0 = 1 0 0 1 1 0 0 0 1 0 0 1 1 1 0 0 0 0 0= 1 1 1 0 0 0 0 0

29. Shift Right operation SHR SHR (shift right) shifts each bit one place to the right. The original rightmost bit is lost and a 0 is shifted into the leftmost position. Ex 1.SHR1 0 1 1 Ex 2.SHR0 0 1 1 = 0 0 0 1 0 1 0 1 1 = 0 1 0 1

30. Shift Right – Multiple Bits SHR # bits means to shift right # times Ex 1: SHR 31 0 0 1 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 = 0 0 0 1 0 0 1 1 Ex 2: SHR 21 1 1 0 0 1 1 0 = 0 0 1 1 1 0 0 1

31. Arithmetic Shift Right operation ASR (retains rightmost sign bit) Shifts each bit one place to the right. The original rightmost bit is lost and a the value of the most significant bit (leftmost bit) is shifted into the new leftmost position. Ex 1.ASR1 0 1 1 Ex 2.ASR0 0 1 1 = 0 0 0 1 1 1 0 1 1= 1 1 0 1

32. Arithmetic Shift Right – Multiple Bits ASR # bits means to arithmetic shift right # times Ex 1: ASR 31 0 0 1 1 1 0 0 1 1 1 1 0 0 1 1 1 0 0 = 1 1 1 1 0 0 1 1 Ex 2: ASR 20 1 1 0 0 1 1 0 = 0 0 0 1 1 0 0 1

33. Rotate Left operation ROL ROL (rotate left) shifts each bit one place to the left. The original leftmost bit is shifted into the rightmost position. No bits are lost. Ex 1.ROL1 1 0 1 Ex 2. ROL1 1 0 0 = 1 0 0 1 1 0 11

34. Rotate Left – Multiple Bits ROL # bits means to rotate left # times Ex 1: ROL 31 0 0 1 1 1 0 0 = 1 1 1 0 0 1 0 0 Ex 2: ROL 21 1 1 0 0 1 1 0 = 1 0 0 1 1 0 1 1

35. Rotate Right operation ROR ROR (rotate right) shifts each bit one place to the right. The original rightmost bit is shifted into the leftmost position. No bits are lost. Ex 1.ROR1 0 1 1 Ex 2.ROR0 0 1 1 = 1 0 0 1 1 0 11

36. Rotate Right – Multiple Bits ROR # bits means to rotate right # times Ex 1: ROR 31 0 0 1 1 1 0 0 = 1 0 0 1 0 0 1 1 Ex 2: ROR 21 1 1 0 0 1 1 0 = 1 0 1 1 1 0 0 1