Bit Manipulation and Masking

7/23/2019 Bit Manipulation and Masking

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Bit Manipulation in 'C' 'C' bit operators

OP symbol Description Example& AND y = y & 0xF7;

| OR y = y | 0x08;^ XOR>> n shift bits right n places y = y >> 4;<< n shift bits left n places y = y << 2;

~ complements (invert all bits) y = ~y;

Only use bit operators with integer data types - Not real data types.

Note: In 'C' binary operators(such as y = y OP z;) can be written as in a

shortened form: y OP= z;E.g.'s y = y + 7; y+=7;

y = y & 0x80; y &= 0x80;y = y >> 4; y >>= 4;



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Masking for bit manipulation A mask is a pattern of 0's and 1's

Use OR ('|')and a mask value to set selected bits to '1'

Use AND ('&') and a mask value to set selected bits to '0' Use XOR(^) to toggle bits (0 -> 1 and 1-> 0)

OR

A B Output

0 0 0

0 1 11 0 1

1 1 1

AND

A B Output

0 0 0

0 1 01 0 0

1 1 1

XOR

A B Output

0 0 0

0 1 11 0 1

1 1 0



Shift Operations << shift left

used to move bits to the left - lose MSB(mostsignificant bit) and gain new LSB(least significant bit)New LSB = 0 for all data types.

can be used for multiply by 2 for each shift.

>> shift right

used to move bits to the right - lose LSB and gainnew MSB (Note: new MSB = 0 for unsigned integer

data types, or previous MSB for signed types).

can be used for divide by 2 for each shift.

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Bit Masking - Setting bits to 1 Setting bits to 1 using the OR (|) bit-wise operator

General format: value = value | mask; Shorthand : value |= mask; Mask : '1' to set a bit, '0' to leave bit unchanged.

Example: Controlling a single bit (Note var is an integer data type.)

To set bit 5 to '1' the mask is 0000….0000100000var= var | 0x20; //mask expressed as hexadecimal value

or shortended formvar |= 0x20;

Is often written using the shift left operator :-

var |= (1UL << 5); // note the brackets to force shift before ORing

1UL - UL represents unsigned long i.e. 32 bit representation of 1<< 5 - shift bit pattern 5 places to the left.

- result is 000….00100000 or 0x00000020



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Bit Masking - Setting bits to 0 Setting bits to 0 using the AND (&) bit-wise operator

General format: value = value & mask; Shorthand : value &= mask; Mask : '0' to clear a bit, '1' to leave bit unchanged

Example: Controlling a single bit (Note var is an integer data type.)

To set bit 5 to '0'For a 32bit integer the mask is 11111111111111111111111111011111

var = var & 0xFFFFFFDF; or var &= 0xFFFFFFDF;

or

var &= ~0x20;or

var &= ~(1UL << 5);



To set and clear multiple bits Set bits 1 and 4 of var

var |= ((1UL << 1) | (1 UL << 4));

Clear bits 1 and 17 of var

var &= ~( (1UL << 1) | (1UL << 17) );

what do these do?

1. var |= ((1UL << 0) | (1UL << 5) | (1UL << 12) );

2. var &= ~((1UL << 0) | (1UL << 5) | (1UL << 12) );

3. var |= (3UL << 6);

4. var &= ~ (3UL << 2 * 3);

5. var |= (1UL << 6) | (1UL << 7) | (1UL << 8);

6. var &= ~(3UL << 2*10);

7. var |= (1UL << 2*10);

var &= ~((3UL << 2*2) | (3UL << 2*3) | (3UL << 2*6) | (3UL << 2*7));

var |= ((1UL << 2*2) | (1UL << 2*3) | (1UL << 2*6) | (1UL << 2*7));

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Some for you to try Write expressions to perform the following

operations on a variable x.1. set bit 11

2. clear bit 13

3.set bits 6 and 10

4. clear bits 13 and 14

5. set bits 4,5,6, and 7

6.set bit 2 and clear bit 3

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Input Masking Used to test the value of a single bit.

Use AND (&) and a mask to isolate the selected bit

value & mask Used in input Polling to test bit status set bit(s) to be tested to '1', all other bits to '0'Example: Testing bit 3

mask is 00001000 (0x08) or (1UL << 3) Two possible results 00000000 or 00001000 when ANDED I.e. zero or non-zero

Loop while bit is '0', exit when bit becomes 1

while ( (reg & (1UL << 3)) == 0) { ; } Loop while bit is '1', exit when bit becomes 0

while ( (reg & (1UL << 3)) != 0) { ; }



Testing multiple bits use a mask with each bit to be tested set to 1

test for bits 3 and 7 both at logic 0 if ( (reg & (1<<3))==0) && (reg & (1<<7) == 0) ) {

test for bits 4 and 6 both at logic 1

if ( (reg & (1<<4))!=0) && (reg & (1<<6) == 0) ) {

test if either bit 3 or bit 7 at logic 0

if ( (reg & (1<<3))==0) || (reg & (1<<7) == 0) ) {

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An Exercise A32 bit register(REG) is interfaced to the following peripherals:-

A motor connected to bit 7 (0 is off, 1 is on)

An led connected to bit 0 (0 is off, 1 is on)

Two switches S1 and S2 connected to bits 1 and 2 respectively thatgenerate either logic 0 or 1 inputs.

Assume inputs start at logic 0 and all output are initialised to 0.

Write C code to perform the following steps:

1. loop waiting while switch S1 is at logic 0, end loop when S1 becomes 1.2. Now turn on the led

3. loop waiting until both switches are at logic 1

4. Now turn on the motor

5. Loop waiting until either of the switches change to logic 06. Turn of the motor and the led

7. loop until both switches are at logic 0

8. start back at step 1

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Bit Manipulation and Masking