- Recall how to
Set a certain bit
in variable:
a = 1010...x...1101 0000..010..0000 <------------- Or this bit pattern OR --------------- 1010...1...1101We can best use the base 2, 8 or 16 number notation to denote bit patterns !!!
- Example:
set
the 2nd bit
and the 4th bit in
a bit pattern
int main( int argc, char* argv[] ) { char a = 1; /* 1 = 00000001 */ a = a | 0b00000100; /* 00000100 - set 2nd bit */ a = a | 0b00010000; /* 00010000 - set 4th bit */ printf( "a = %d\n", a ); }
- Clear a certain bit in a
bit pattern
- Suppose the variable a
contains a bit pattern:
a = 1010...x...1101
- Problem:
- Clear the bit x to the value 1 without changing all other bits in the pattern.
- We can clear the bit
x inside the
bit pattern to 0 using
the AND operation with
the following
specific pattern:
a = 1010...x...1101 1111..101..1111 AND --------------- 1010...0...1101
Note:
- The bit pattern
1111...101...1111can be obtained using:
NOT ( 0000...010...0000 )
- Suppose the variable a
contains a bit pattern:
- Example:
clear
the 4th bit
and the 2nd bit in
a bit pattern
int main( int argc, char* argv[] ) { char a = 31; /* a = 00011111 (= 31) */ a = a & (~0b00010000); /* Now: a = 00001111 (= 15) */ printf( "a = %d\n", a ); a = a & (~0b0000100); /* Now: a = 00001011 (= 11) */ printf( "a = %d\n", a ); }
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link and
save in a scratch directory
- To compile: gcc bit-op3.c
- To run: ./a.out
- Flipping a bit:
- Flipping a bit = changing a 0 bit to 1 or changing a 1 bit to 0
- Flip a certain bit in a
bit pattern
- Suppose the variable a
contains a bit pattern:
a = 1010...x...1101
- Problem:
- Flip the bit x without changing all other bits in the pattern.
- We can flip the bit
x inside the
bit pattern using
the XOR operation with
the following
specific pattern:
a = 1010...x...1101 0000..010..0000 XOR --------------- 1010...x...1101
- Suppose the variable a
contains a bit pattern:
- Example:
flip
the 4th bit
in
a bit pattern OFF and ON
int main( int argc, char* argv[] ) { char a = 31; /* a = 00011111 (= 31) */ a = a ^ 0b00010000; /* Now: a = 00001111 (= 15) */ printf( "a = %d\n", a ); a = a ^ 0b00010000; /* Now: a = 00011111 (= 31) */ printf( "a = %d\n", a ); }
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link and
save in a scratch directory
- To compile: gcc bit-op5.c
- To run: ./a.out
- Testing a certain bit in a
bit pattern
- Suppose the variable a
contains a bit pattern:
a = 1010...x...1101
- Problem:
- Test whether the bit x is 1 or not without changing any bit in the pattern.
- We can test the bit
x inside the
bit pattern by
first
filtering out
this bit using
the AND operation with
the following
specific pattern:
a = 1010...x...1101 0000..010..0000 AND --------------- 0000...x...0000and then:
- Testing whether the result is equal to 0 or not.
- Suppose the variable a
contains a bit pattern:
- Example: test
whether the 3rd bit in
a bit pattern is set or reset
int main( int argc, char* argv[] ) { char a = .....; /* Any bit pattern */ if ( (a & 0b00001000) == 0 ) /* Test is 3rd bit is set */ printf( "bit is 0\n" ); else printf( "bit is 1\n" ); }
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link and
save in a scratch directory
- To compile: gcc bit-op4.c
- To run: ./a.out
- We will see soon that there is a better way to construct bit patterns that are used in the bit manipulation operations
- These bit wise operators are
also defined in Java.
- They are often omitted in introduction courses because they are not used in program at the introduction level.