Binary operations

Binary Operations
Math/Logical

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

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.

Binary Addition Rules
Rules:
 0 + 0 = 0
 0 + 1 = 1
 1 + 0 = 1 (just like in decimal)
 1 + 1 = 210
= 102 = 0 with 1 to carry
 1 + 1 + 1 = 310
= 112 = 1 with 1 to carry

Binary Addition Example 1
1 1 0 1 1 1
+ 0 1 1 1 0 0
1
1
111
0 1 0 0 11
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

Binary Addition Explanation
1 1 0 1 1 1
+ 0 1 1 1 0 0
- .
1
1
111
0 1 0 0 11
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
In column 7, we just bring down the
carried 1
2
2 2 23
222
What is actually
happened when we
carried in binary?

Binary Addition Verification
Verification
1101112  5510
+0111002 + 2810
8310
64 32 16 8 4 2 1
1 0 1 0 0 1 1
= 64 + 16 + 2 +1
= 8310
1 1 0 1 1 1
+ 0 1 1 1 0 0
1 0 1 0 0 11
You can always check your
answer by converting the
figures to decimal, doing the
addition, and comparing the
answers.

Binary Addition Example 2
Verification
1110102  5810
+0011112 + 1510
7310
64 32 16 8 4 2 1
1 0 0 1 0 0 1
= 64 + 8 +1
= 7310
1 1 1 0 1 0
+ 0 0 1 1 1 1
1
1
11
0 0 1 0 10
Example 2:
Add 1111 to 111010.
11

Decimal Subtraction
Example
8 0 2 5
- 4 6 5 7
Subtract
4657 from 8025:
7 9 11
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.
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.
Then subtract 15 – 7 = 8.
Add the borrowed 10 to the original 1..
Then subract 11 – 5 = 6

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

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.

Binary Subtraction
Example 1
1 1 0 0 1 1
- 1 1 1 0 0
Example 1: Subtract
binary 11100 from 110011
2
0 0 2
1
2
1101
Col 1) Subtract 1 – 0 = 1
Col 5) Try to subtract 0 – 1  can’t.
Must borrow from next column.
Col 3) Try to subtract 0 – 1  can’t.
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).
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.

Binary Subtraction
Verification
Verification
1100112  5110
- 111002 - 2810
2310
64 32 16 8 4 2 1
1 0 1 1 1
= 16 + 4 + 2 + 1
= 2310
1 1 0 0 1 1
- 1 1 1 0 0
2
0 0 2
1
2
1101 1
Subtract binary
11100 from 110011:

Binary Subtraction
Example 2
1 0 1 0 0 1
- 1 0 1 0 0
Example 2: Subtract
binary 10100 from 101001
20 0 2
1101 0
Verification
1010012  4110
- 101002 - 2010
2110
64 32 16 8 4 2 1
1 0 1 0 1
= 16 + 4 + 1
= 2110

Logical, Shift, and Rotate
Operations

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.

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

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
OR 0 0 0 0 1 1 1 1
1 0 0 1 1 1 1 1

Logical OR Examples
Ex 3: 0 1 1 1
OR 0 0 1 0
0 1 1 1
Ex 2: 1 1 0 0 1 0 0 1
OR 0 0 0 0 1 0 1 0
1 1 0 0 1 0 1 1

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

Logical XOR Examples
Ex 1: 1 0 0 1 1 0 0 1
XOR 0 0 0 0 1 1 1 1
1 0 0 1 0 1 1 0
Ex 2: 0 1 1 1
XOR 0 0 1 0
0 1 0 1

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

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

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

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.

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. SHL 1 1 0 1
Ex 2. SHL 1 1 0 0
= 1 0 0 0
01 1 0 11 1 0 1 = 1 0 1 0

Shift Left - Multiple Bits
 SHL # bits means to shift left # times
Ex 1: SHL 3 1 0 0 1 1 1 0 0
Ex 2: SHL 2 1 1 1 0 0 1 1 0
= 1 0 0 1 1 0 0 0
1 0 0 1 1 1 0 01 0 0 1 1 1 0 0 0 0 0 = 1 1 1 0 0 0 0 0

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. SHR 1 0 1 1
Ex 2. SHR 0 0 1 1
= 0 0 0 1
0 1 0 1 1 = 0 1 0 1

Shift Right – Multiple Bits
 SHR # bits means to shift right # times
Ex 1: SHR 3 1 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 2 1 1 1 0 0 1 1 0
= 0 0 1 1 1 0 0 1

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. ASR 1 0 1 1
Ex 2. ASR 0 0 1 1
= 0 0 0 1
1 1 0 1 1 = 1 1 0 1

Arithmetic Shift Right –
Multiple Bits
 ASR # bits means to arithmetic shift right # times
Ex 1: ASR 3 1 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 2 0 1 1 0 0 1 1 0
= 0 0 0 1 1 0 0 1

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. ROL 1 1 0 1
Ex 2. ROL 1 1 0 0
= 1 0 0 1
1 0 1 1

Rotate Left – Multiple Bits
 ROL # bits means to rotate left # times
Ex 1: ROL 3 1 0 0 1 1 1 0 0
= 1 1 1 0 0 1 0 0
Ex 2: ROL 2 1 1 1 0 0 1 1 0
= 1 0 0 1 1 0 1 1

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. ROR 1 0 1 1
Ex 2. ROR 0 0 1 1
= 1 0 0 1
1 0 11

Rotate Right – Multiple Bits
 ROR # bits means to rotate right # times
Ex 1: ROR 3 1 0 0 1 1 1 0 0
= 1 0 0 1 0 0 1 1
Ex 2: ROR 2 1 1 1 0 0 1 1 0
= 1 0 1 1 1 0 0 1

Binary operations

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