Review: How to set the bit at position k to 1

The SETBIT(x, k) operation: (sets the bit at position k in x)

position k | V Before SETBIT(x,k): x = 1010...?...1101 After SETBIT(x,k): x = 1010...1...1101

Solution: initially, the bit at position k has the value ? (the value can be 0 or 1)

x = 1010...?...1101 pattern = 0000...1...0000 ---------------- OR 1010...1...1101

Review: How to set the bit at position k to 1

The SETBIT(x, k) operation: (sets the bit at position k in x)

position k | V Before SETBIT(x,k): x = 1010...?...1101 After SETBIT(x,k): x = 1010...1...1101

Solution: construct the pattern 000..00100..000 (it has a 1 at position k):

x = 1010...?...1101 pattern = 000..00100..000 ---------------- OR 1010...1...1101

Review: How to set the bit at position k to 1

The SETBIT(x, k) operation: (sets the bit at position k in x)

position k | V Before SETBIT(x,k): x = 1010...?...1101 After SETBIT(x,k): x = 1010...1...1101

Solution: apply an OR operation to set the bit at position k:

x = 1010...?...1101 pattern = 000..00100..000 ---------------- OR x = 1010...1...1101 (y OR 0 = y and ? OR 1 = 1) SETBIT(x,k) ≡ x = x | pattern

The bit at position k is set because: ? OR 1 = 1
The other bits are unchanged because: y OR 0 = y

DEMO: demo/C/set1/bit-op2a.c

How to clear the bit at position k to 1

The CLRBIT(x, k) operation: (clears the bit at position k in x)

position k | V Before CLRBIT(x,k): x = 1010...?...1101 After CLRBIT(x,k): x = 1010...0...1101

Solution: initially, the bit at position k has the value ? (the value can be 0 or 1)

x = 1010...?...1101 pattern = 0000...1...0000 ---------------- OR 1010...1...1101

How to clear the bit at position k to 1

The CLRBIT(x, k) operation: (clears the bit at position k in x)

position k | V Before CLRBIT(x,k): x = 1010...?...1101 After CLRBIT(x,k): x = 1010...0...1101

Solution: construct the pattern 111..11011..111 (it has a 0 at position k):

x = 1010...?...1101 pattern = 111..11011..111 ---------------- OR 1010...1...1101

How to clear the bit at position k to 1

The CLRBIT(x, k) operation: (clears the bit at position k in x)

position k | V Before CLRBIT(x,k): x = 1010...?...1101 After CLRBIT(x,k): x = 1010...0...1101

Solution: apply an AND operation to clear the bit at position k:

x = 1010...?...1101 pattern = 111..11011..111 ---------------- AND x = 1010...0...1101 (? AND 0 = 0 and y AND 1 = y) CLRBIT(x,k) ≡ x = x & pattern

The bit at position k is cleared because: ? AND 0 = 0
The other bits are unchanged because: y AND 1 = y

How to obtain the pattern used in the CLRBIT(x,k) operation

The pattern used to clear a bit at position k is:

position k | V pattern = 111..11011..111

It's easier to construct the pattern using the pattern 000..00100..000:

111..11011..111 ≡ ~ 000..00100..000 ^ | the bitwise NOT operation (flip each bit) !

DEMO: demo/C/set1/bit-op3.c

How to flip the bit at position k

The FLIPBIT(x, k) operation: (flips the bit at position k in x)

position k | V Before FLIPBIT(x,k): x = 1010...?...1101 After FLIPBIT(x,k): x = 1010...¿...1101

Solution: initially, the bit at position k has the value ? (the value can be 0 or 1)

x = 1010...?...1101 pattern = 0000...1...0000 ---------------- OR 1010...1...1101

How to flip the bit at position k

The FLIPBIT(x, k) operation: (flips the bit at position k in x)

position k | V Before FLIPBIT(x,k): x = 1010...?...1101 After FLIPBIT(x,k): x = 1010...¿...1101

Solution: construct the pattern 000..00100..000 (it has a 1 at position k):

x = 1010...?...1101 pattern = 000..00100..000 ---------------- OR 1010...1...1101

How to flip the bit at position k

The FLIPBIT(x, k) operation: (flips the bit at position k in x)

position k | V Before FLIPBIT(x,k): x = 1010...?...1101 After FLIPBIT(x,k): x = 1010...¿...1101

Solution: apply an XOR operation to flip the bit at position k:

x = 1010...?...1101 pattern = 000..00100..000 ---------------- XOR x = 1010...¿...1101 (y XOR 0 = y and ? XOR 1 = ¿) FLIPBIT(x,k) ≡ x = x ^ pattern

The bit at position k is flipped because: ? XOR 1 = ¿ (0 XOR 1 = 1 and 1 XOR 1 = 0)
The other bits are unchanged because: y XOR 0 = y

DEMO: demo/C/set1/bit-op5.c

How to test if the bit at position k is 0 or 1

The ISSET(x, k) operation: (returns true (!= 0) if bit k in x is 1, and returns false (== 0) otherwise)

position k | if ? == 0: ISSET(x,k) == 0 V if ? == 1: ISSET(x,k) != 0 x = 1010...?...1101

Solution: initially, the bit at position k has the value ? (the value can be 0 or 1)

x = 1010...?...1101 pattern = 0000...1...0000 ---------------- OR 1010...1...1101

How to test if the bit at position k is 0 or 1

The ISSET(x, k) operation: (returns true (!= 0) if bit k in x is 1, and returns false (== 0) otherwise)

position k | if ? == 0: ISSET(x,k) == 0 V if ? == 1: ISSET(x,k) != 0 x = 1010...?...1101

Solution: construct the pattern 000..00100..000 (it has a 1 at position k):

x = 1010...?...1101 pattern = 000..00100..000 ---------------- OR 1010...1...1101

How to test if the bit at position k is 0 or 1

The ISSET(x, k) operation: (returns true (!= 0) if bit k in x is 1, and returns false (== 0) otherwise)

position k | if ? == 0: ISSET(x,k) == 0 V if ? == 1: ISSET(x,k) != 0 x = 1010...?...1101

Solution: apply an AND operation to test the bit at position k:

x = 1010...?...1101 pattern = 000..00100..000 ---------------- AND x = 0000...?...0000 (y AND 0 = 0 and ? AND 1 = ?) ISSET(x,k) ≡ (x & pattern) != 0

The bit at position k is retained because: ? AND 1 = ?
All the other bits become 0 bits because: y AND 0 = 0

DEMO: demo/C/set1/bit-op4.c