DevTut

# Bit Operators

# | - bitwise OR

int a = 5;     // 0101b  (0x05)
int b = 12;    // 1100b  (0x0C)
int c = a | b; // 1101b  (0x0D)

std::cout << "a = " << a << ", b = " << b << ", c = " << c << std::endl;

Output

a = 5, b = 12, c = 13

Why

A bit wise OR operates on the bit level and uses the following Boolean truth table:

true OR true = true
true OR false = true
false OR false = false

When the binary value for a (0101) and the binary value for b (1100) are OR'ed together we get the binary value of 1101:

int a = 0 1 0 1
int b = 1 1 0 0 |
        ---------
int c = 1 1 0 1

The bit wise OR does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator |=:

int a = 5;  // 0101b  (0x05)
a |= 12;    // a = 0101b | 1101b

# ^ - bitwise XOR (exclusive OR)

int a = 5;     // 0101b  (0x05)
int b = 9;     // 1001b  (0x09)
int c = a ^ b; // 1100b  (0x0C)

std::cout << "a = " << a << ", b = " << b << ", c = " << c << std::endl;

Output

a = 5, b = 9, c = 12

Why

A bit wise XOR (exclusive or) operates on the bit level and uses the following Boolean truth table:

true OR true = false
true OR false = true
false OR false = false

Notice that with an XOR operation true OR true = false where as with operations true AND/OR true = true, hence the exclusive nature of the XOR operation.

Using this, when the binary value for a (0101) and the binary value for b (1001) are XOR'ed together we get the binary value of 1100:

int a = 0 1 0 1
int b = 1 0 0 1 ^
        ---------
int c = 1 1 0 0

The bit wise XOR does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator ^=:

int a = 5;  // 0101b  (0x05)
a ^= 9;    // a = 0101b ^ 1001b

The bit wise XOR can be utilized in many ways and is often utilized in bit mask operations for encryption and compression.

Note: The following example is often shown as an example of a nice trick. But should not be used in production code (there are better ways std::swap() to achieve the same result).

You can also utilize an XOR operation to swap two variables without a temporary:

int a = 42;
int b = 64;

// XOR swap
a ^= b;
b ^= a;
a ^= b;

std::cout << "a = " << a << ", b = " << b << "\n";

To productionalize this you need to add a check to make sure it can be used.

void doXORSwap(int& a, int& b)
{
    // Need to add a check to make sure you are not swapping the same
    // variable with itself. Otherwise it will zero the value.
    if (&a != &b)
    {
        // XOR swap
        a ^= b;
        b ^= a;
        a ^= b;
    }
}

So though it looks like a nice trick in isolation it is not useful in real code. xor is not a base logical operation,but a combination of others: a^c=~(a&c)&(a|c)

also in 2015+ compilers variables may be assigned as binary:

int cn=0b0111;

# & - bitwise AND

int a = 6;     // 0110b  (0x06)
int b = 10;    // 1010b  (0x0A)
int c = a & b; // 0010b  (0x02)

std::cout << "a = " << a << ", b = " << b << ", c = " << c << std::endl;

Output

a = 6, b = 10, c = 2

Why

A bit wise AND operates on the bit level and uses the following Boolean truth table:

TRUE  AND TRUE  = TRUE
TRUE  AND FALSE = FALSE
FALSE AND FALSE = FALSE

When the binary value for a (0110) and the binary value for b (1010) are AND'ed together we get the binary value of 0010:

int a = 0 1 1 0
int b = 1 0 1 0 &
        ---------
int c = 0 0 1 0

The bit wise AND does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator &=:

int a = 5;  // 0101b  (0x05)
a &= 10;    // a = 0101b & 1010b

# << - left shift

int a = 1;      // 0001b
int b = a << 1; // 0010b

std::cout << "a = " << a << ", b = " << b << std::endl;

Output

a = 1, b = 2

Why

The left bit wise shift will shift the bits of the left hand value (a) the number specified on the right (1), essentially padding the least significant bits with 0's, so shifting the value of 5 (binary 0000 0101) to the left 4 times (e.g. 5 << 4) will yield the value of 80 (binary 0101 0000). You might note that shifting a value to the left 1 time is also the same as multiplying the value by 2, example:

int a = 7;
while (a < 200) {
    std::cout << "a = " << a << std::endl;
    a <<= 1;
}

a = 7;
while (a < 200) {
    std::cout << "a = " << a << std::endl;
    a *= 2;
}

But it should be noted that the left shift operation will shift all bits to the left, including the sign bit, example:

int a = 2147483647; // 0111 1111 1111 1111 1111 1111 1111 1111
int b = a << 1;     // 1111 1111 1111 1111 1111 1111 1111 1110

std::cout << "a = " << a << ", b = " << b << std::endl;

Possible output: a = 2147483647, b = -2

While some compilers will yield results that seem expected, it should be noted that if you left shift a signed number so that the sign bit is affected, the result is undefined. It is also undefined if the number of bits you wish to shift by is a negative number or is larger than the number of bits the type on the left can hold, example:

int a = 1;
int b = a << -1;  // undefined behavior
char c = a << 20; // undefined behavior

The bit wise left shift does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator <<=:

int a = 5;  // 0101b
a <<= 1;    // a = a << 1;

# >> - right shift

int a = 2;      // 0010b
int b = a >> 1; // 0001b

std::cout << "a = " << a << ", b = " << b << std::endl;

Output

a = 2, b = 1

Why

The right bit wise shift will shift the bits of the left hand value (a) the number specified on the right (1); it should be noted that while the operation of a right shift is standard, what happens to the bits of a right shift on a signed negative number is implementation defined and thus cannot be guaranteed to be portable, example:

int a = -2;    
int b = a >> 1; // the value of b will be depend on the compiler

It is also undefined if the number of bits you wish to shift by is a negative number, example:

int a = 1;
int b = a >> -1;  // undefined behavior

The bit wise right shift does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator >>=:

int a = 2;  // 0010b
a >>= 1;    // a = a >> 1;

# ~ - bitwise NOT (unary complement)

unsigned char a = 234;  // 1110 1010b  (0xEA)
unsigned char b = ~a;   // 0001 0101b  (0x15)

std::cout << "a = " << static_cast<int>(a) <<
             ", b = " << static_cast<int>(b) << std::endl;

Output

a = 234, b = 21

Why

A bit wise NOT (unary complement) operates on the bit level and simply flips each bit. If it's a 1, it's changed to a 0, if it's a 0, it's changed to a 1. The bit wise NOT has the same effect as XOR'ing a value against the max value for a specific type:

unsigned char a = 234;  // 1110 1010b  (0xEA)
unsigned char b = ~a;   // 0001 0101b  (0x15)
unsigned char c = a ^ ~0;

The bit wise NOT can also be a convenient way to check the maximum value for a specific integral type:

unsigned int i = ~0;
unsigned char c = ~0;

std::cout << "max uint = " << i << std::endl <<
             "max uchar = " << static_cast<short>(c) << std::endl;

The bit wise NOT does not change the value of the original value and does not have a compound assignment operator, so you can not do a ~= 10 for example.

The bit wise NOT (~) should not be confused with the logical NOT (!); where a bit wise NOT will flip each bit, a logical NOT will use the whole value to do its operation on, in other words (!1) != (~1)

# Remarks

Bit shift operations are not portable across all processor architectures, different processors can have different bit-widths. In other words, if you wrote

int a = ~0;
int b = a << 1;

This value would be different on a 64 bit machine vs. on a 32 bit machine, or from an x86 based processor to a PIC based processor.

Endian-ness does not need to be taken into account for the bit wise operations themselves, that is, the right shift (>>) will shift the bits towards the least significant bit and an XOR will perform an exclusive or on the bits. Endian-ness only needs to be taken into account with the data itself, that is, if endian-ness is a concern for your application, it's a concern regardless of bit wise operations.

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