CSE 1320: INTERMEDIATE PROGRAMMING BITWISE...

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CSE 1320: INTERMEDIATE

PROGRAMMING

BITWISE OPERATORS

Mingon Kang

The University of Texas at Arlington

Revisiting printf() function (refer to revisit_printf.c)

%d for Decimal (Base 10)

%x, %X for hexadecimal (Base 16)

%o for octal (Base 8)

Do we have a binary (Base 2) format specifier?

Revisiting printf() function

Lecture 19: Bitwise Operators 2

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N-bit integer system

Most significant bit (left-

most bit) denotes the sign

bit

0 indicates positive

1 indicates negative

2’s Complement for signed representation


Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1


Most significant bit (left-

most bit) denotes the sign

bit

0 indicates positive

1 indicates negative



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1

2′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑥 = −𝑥 Example:

2′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 5 = −5

2′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 −5 = 5

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2’s complement is an

operation of changing

the sign of the given

integer.



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1

2′𝑠 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑥 = 1′𝑠 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑥 + 1

Then what is 1’s Complement of x?


1’s Complement of x:

The number produced after

you subtract each bits of x

from 1

1′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 0112 =1′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 3 = 1002,

where N=3

1′𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 11002 = 00112,

where N=4



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1


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2’s Complement of x:

2′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 0112 == 2′ 𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 3

= 1′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡(0112) + 1

= 1002 + 1 = 1012



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1



2’s Complement of x = -x

One exception to this negativity:

2′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 1002 = 2′ 𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 −4

= 1′𝑠𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡(1002) + 1

= 0112 + 1 = 1002 = −𝟒



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1


So, this negative number is the

minimum in the range of number using N-bits

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The range of representable

integers:

When N=3:

-4 to 3

When N=4:

-8 to 7

When N:

(−2𝑁−1) 𝑡𝑜 (2𝑁−1 − 1)



Binary Decimal

(unsigned)

Decimal

(signed)

000 0 0

001 1 1

010 2 2

011 3 3

100 4 -4

101 5 -3

110 6 -2

111 7 -1


Now you can replace any subtraction, division,

multiplication operation using only the addition

operation !! (A Giant Step towards circuit minimization)



5 – 2 = ?

= 5 + (-2)

(Let, N=4 bits)

2’sComplement(2)= -2 = 1’sComplement(2)+1

= 1’sComplement(00102)+1

= 11012 + 1 = 11102

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Now you can replace any subtraction, division,

multiplication operation using only the addition

operation !! (A Giant Step towards circuit minimization)



5 – 2 = ?

= 5 + (-2)

= 01012 + 11102 = ? (N=4bits) 01012 11102

------

1 00112 = 3

What is Bitwise Operator?

Allows you to get, set, shift bits in an integer type

variable declared in C

Tools (operators) to be used:

Bitwise NOT (~) : Is the 1’s complement operator

Bitwise AND (&):

0 & 0 = 0

0 & 1 = 0

1 & 0 = 0

1 & 1 = 1


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Bitwise AND (&)

Bitwise OR ( | ):

0 | 0 = 0

0 | 1 = 1

1 | 0 = 1

1 | 1 = 1





Bitwise AND (&)

Bitwise OR ( | )

Bitwise XOR (^):

0 ^ 0 = 0

0 ^ 1 = 1

1 ^ 0 = 1

1 ^ 1 = 0


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Bitwise AND (&)

Bitwise OR ( | )

Bitwise XOR (^)

Left Shift Operator (<<):

x = 010011012

x << 1

Value of x = 1001101𝟎2

x << 1

Value of x = 001101𝟎𝟎2

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• 𝑦 = 101101002

• y << 3

• Value of y = 10100𝟎𝟎𝟎2




Bitwise AND (&)

Bitwise OR ( | )

Bitwise XOR (^)

Left Shift Operator (<<)

Right Shift Operator (>>):

x = 010011012

x >> 1

Value of x = 𝟎01001102

x >> 1

Value of x =𝟎𝟎0100112

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• 𝑦 = 101101002

• y >> 3

• Value of y = 𝟏𝟏𝟏101102

(sign extension if y is a signed

integer, although it is

architecture dependent)

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[refer to bitwise_operators_example1.c]

Bitwise NOT (~)

Bitwise AND (&)

Bitwise OR ( | )

Bitwise XOR (^)

Left Shift Operator (<<)

Right Shift Operator (>>)

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Why bother learning it?

Packing of (Boolean) Variables

An N-bit integer variable can pack N Boolean

variables in it

How can you access 3rd from the right variable from the

packed variable?

How can you change of value to the 5th from the right

variable in the packed variable?


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Accessing 3rd bit from the right


𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0

Suppose, N = 8 bit integer system (unsigned)

𝒙 =

? 1 & 𝑥2 = 𝑥2

0 & 𝑥2 = 0

Accessing 3rd bit from the right

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𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

0 0 0 0 0 𝟏 0 0

& & & & & & & &

Result= 0 0 0 0 0 𝑥2 0 0

𝑅𝑒𝑠𝑢𝑙𝑡 = 1 𝑖𝑓 𝑥2 = 10 𝑖𝑓 𝑥2 = 0

0 0 0 0 0 0 0 1 mask = << 2

mask =

if ( (x & (1<<2) ) != 0) printf(“1”);

else printf(“0”);

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Set 5th bit from the Right to 1


𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

1 | 𝑥4 = 1

0 | 𝑥4 = 𝑥4



𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

0 0 0 1 0 0 0 0

| | | | | | | |

Result= 𝑥7 𝑥6 𝑥5 𝟏 𝑥3 𝑥2 𝑥1 𝑥0

0 0 0 0 0 0 0 1 mask = << 4

mask =

x = x | (1 << 4) ;

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𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

1 1 1 0 1 1 1 1

& & & & & & & &

𝑥7 𝑥6 𝑥5 𝟎 𝑥3 𝑥2 𝑥1 𝑥0

0 0 0 0 0 0 0 1 mask = << 4

~ mask =

0 0 0 1 0 0 0 0

x = x & (~(1 << 4) );

Toggle (Complement) 5th bit from the right


𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

1 ^ 𝑥4 = 𝑥4

0 ^ 𝑥4 = 𝑥4

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Toggle (Complement) 5th bit from the right


𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

0 0 0 1 0 0 0 0 mask =

^ ^ ^ ^ ^ ^ ^ ^

𝑥7 𝑥6 𝑥5 𝒙𝟒 𝑥3 𝑥2 𝑥1 𝑥0 R𝐞𝐬𝐮𝐥𝐭 =

x = x ^ (1 << 4);

x = x ^ 0x10;

Alternatively,

What is the integer equivalent value represented

by the bits 𝒙𝟓𝒙𝟒 𝒙𝟑?


𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥1 𝑥0


𝒙 =

0 0 𝟏 𝟏 𝟏 0 0 0 mask =

& & & & & & & &

0 0 𝒙𝟓 𝒙𝟒 𝒙𝟑 0 0 0 𝑾𝒆 𝒈𝒆𝒕 =

We right shift this to align the three bits to the right,

so that 𝒙𝟑 becomes the least-significant bit.

0 0 𝟎 𝟎 𝟎 𝒙𝟓 𝒙𝟒 𝒙𝟑

>> 3

printf(“%d”, (x & 0x38) >> 3 );

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Multiplying x by 2𝑘

𝑥 ≪ 𝑘


Divide x by 2𝑘

𝑥 ≫ 𝑘


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Check if x is an integer power of 2


if( (x & (x-1)) == 0 )

printf(“YES!!”);

else

printf(“NO”);

x = 4

00100

00011

-------

00000 == 0 (YES)

x = 8

01000

00111

--------

00000 == 0 (YES)

x = 12

01100

01011

-------

01000 != 0 (No)

Even or ODD?


if( x & 1 == 1 )

printf(“ODD!!”);

else

printf(“EVEN”);

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Multi-options as a single parameter

In Windows Programming,

Multiple options can

be passed as a single

parameter by using

bitwise OR operator


HWND WINAPI CreateWindowEx(

_In_ DWORD dwExStyle,

_In_opt_ LPCTSTR lpClassName,

_In_opt_ LPCTSTR lpWindowName,

_In_ DWORD dwStyle,

_In_ int x,

_In_ int y,

_In_ int nWidth,

_In_ int nHeight,

_In_opt_ HWND hWndParent,

_In_opt_ HMENU hMenu,

_In_opt_ HINSTANCE hInstance,

_In_opt_ LPVOID lpParam

);

Multi-options as a single parameter

WS_EX_APPWINDOW| WS_EX_CLIENTEDGE |

WS_EX_RIGHTSCROLLBAR| ..

https://msdn.microsoft.com/en-

us/library/windows/desktop/ms632680(v=vs.85).

aspx


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References

Bitwise Operators in C

(http://en.wikipedia.org/wiki/Bitwise_operations_i

n_C )


http://en.wikipedia.org/wiki/Bitwise_operations_in_C




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