# Bitwise Operations

CS 301: Assembly Language Programming Lecture, Dr. Lawlor

There are a whole group of "bitwise" operators that operate on bits.
 Name C++ x86 Useful to... AND & (ampersand) and mask out bits (set other bits to zero) OR |   (pipe) or reassemble bit fields XOR ^ (caret) xor invert selected bits NOT ~ (tilde) not invert all the bits in a number Left shift << shl makes numbers bigger by shifting their bits to higher places Right shift >> shr sar SHift Right: makes numbers smaller by shifting their bits to lower places. Shift Arithmetic Right: signed shift retains sign bit for negative numbers.

If you'd like to see the bits inside a number, you can loop over the bits and use AND to extract each bit:

```int i=9; // 9 == 8 + 1 == 1001

for (long bit=31;bit>=0;bit--) { // print each bit
long mask=(1L<<bit); // only this bit is set
long biti=mask&i; // extract this bit from i
if (biti!=0) std::cout<<"1";
else         std::cout<<"0";
}std::cout<<" integer\n";```

(Try this in NetRun now!)

Because binary is hard to read (was that 1000000000000000 or 10000000000000000?), we normally use hexadecimal, base 16.

 Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 Binary 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000

Remember that every hex digit represents four bits.  So if you shift a hex constant by four bits, it shifts by one entire hex digit:

0xf0d<<4 == 0xf0d0
0xf0d>>4 == 0xf0

If you shift a hex constant by a non-multiple of four bits, you end up interleaving the hex digits of the constant, which is confusing:

0xf0>>2 == 0x3C (?)

Bitwise operators make more sense working with hex digits, because they operate on the underlying bits of those digits:
0xff0 & 0x0ff == 0x0f0
0xff0 | 0x0ff == 0xfff
0xff0 ^ 0x0ff == 0xf0f

You can use these bitwise operators to peel off the hex digits of a number, to print out stuff in hex.

`int v=1024+15;for (int digit=7;digit>=0;digit--) {	char *digitTable="0123456789abcdef";	int d=(v>>(digit*4))&0xF;	std::cout<<digitTable[d];}std::cout<<std::endl;return v;`

(Try this in NetRun now!)

You could also use printf("%X",v);

## Bitwise Left Shift: <<

Makes values bigger, by shifting the value's bits into higher places, tacking on zeros in the vacated lower places.

 As Ints As Bits 3<<0 == 3 0011<<0 == 0011 3<<1 == 6 0011<<1 == 0110 3<<2 == 12 0011<<2 == 1100

• 1<<n pushes a 1 up into bit number n, creating the bit pattern 1 followed by n zeros.
• The value of (k<<n) is actually k*2n.  This means bit shifting can be used as a faster multiply by a power of two.
• k<<0 == k, for any k.
• (k<<n) >= k, for any n and k (unless you have "overflow"!).
• If the shift count n is negative, or greater than the number of bits in the value being shifted, C++ allows basically anything to happen (the machine could crash, or catch fire, or return garbage).  In x86 assembly, the shift instruction wraps the count around by the number of bits in the datatype, so (x<<66) == (x<<2).
• Left shift always shifts in fresh new zero bits.
• You can left shift by as many bits as you want.

In C++, the << operator is also overloaded for iostream output.  This was a confusing choice, in particular because "cout<<3<<0;" just prints 3, then 0!  To actually print the value of "3<<0", you need parenthesis, like this: "cout<<(3<<0);".  Operator precedence is screwy for bitwise operators, so you really want to use excess parenthesis!

In assembly:

• shl is "shift left".  Use it like "shl eax,4" (Try this in NetRun now!).  Note that the number of bits to shift can be a constant (here, the 4), or register cl (low bits of ecx), but not any other register (Try this in NetRun now!).
• sal is the same instruction (same machine code).
• There's also a "rol" that does a circular left shift: the bits leaving the left side come back in the right side.

## Bitwise Right Shift: >>

Makes values smaller, by shifting them into lower-valued places.  Note the bits in the lowest places just "fall off the end" and vanish.

 As Ints As Bits 3>>0 == 3 0011>>0 == 0011 3>>1 == 1 0011>>1 == 0001 3>>2 == 0 0011>>2 == 0000 6>>1 == 3 0110>>1 == 0011

• The value of (k>>n) is actually k/2n.  This can be used as a faster divide.
• (k<<n)>>n == k, unless overflow has happened.
• There are two flavors of right shift: signed, and unsigned.  Unsigned shift fills in the new bits with zeros.  Signed shift fills the new bits with copies of the sign bit, so negative numbers stay negative after the shift.

If you're dyslexic, like me, the left shift << and right shift >> can be really tricky to tell apart.  I always remember it like this:

• k<<n  pumps up the value of k (the point of the << is injecting bigness into k)
• k>>n  drains away the value of k (the point of the >> is draining bigness from k)

In assembly:

• shr is the unsigned shift.
• sar is the signed (or "arithmetic") shift.
• Again, there's a circular right shift "ror".

## Bitwise AND: &

Output bits are 1 only if both corresponding input bits are 1.  This is useful to "mask out" bits you don't want, by ANDing them with zero.

 As Ints As Bits 3&5 == 1 0011&0101 == 0001 3&6 == 2 0011&0110 == 0010 3&4 == 0 0011&0100 == 0000

Properties:

• 0=A&0     (AND by 0's creates 0's--used for masking)
• A=A&~0 (AND by 1's has no effect)
• A=A&A    (AND by yourself has no effect)

Bitwise AND is a really really useful tool for extracting bits from a number--you often create a "mask" value with 1's marking the bits you want, and AND by the mask.  For example, this code figures out if bit 2 of an integer is set:
int mask=(1<<2); // in binary: 100
int value=...;           // in binary: xyz
if (0!=(mask&value))  // in binary: x00
...

In C/C++, bitwise AND has the wrong precedence--leaving out the parenthesis in the comparison above gives the wrong answer!  Be sure to use extra parenthesis!

In assembly, it's the "and" instruction.  Very simple!

## Bitwise OR: |

Output bits are 1 if either input bit is 1.  E.g., 3|5 == 7; or 011 | 101 == 111.

 As Ints As Bits 3|0 == 3 0011|0000 == 0011 3|3 == 3 0011|0011 == 0011 1|4 == 5 0001|0100 == 0101

• A=A|0      (OR by 0's has no effect)
• ~0=A|~0 (OR by 1's creates 1's)
• A=A|A      (OR by yourself has no effect)

Bitwise OR is useful for sticking together bit fields you've prepared separately.  Overall, you use AND to pick apart an integer's values, XOR and NOT to manipulate them, and finally OR to assemble them back together.

## Bitwise XOR: ^

Output bits are 1 if either input bit is 1, but not both. E.g., 3^5 == 6; or 011 ^ 101 == 110.  Note how the low bit is 0, because both input bits are 1.

 As Ints As Bits 3^5 == 6 0011&0101 == 0110 3^6 == 5 0011&0110 == 0101 3^4 == 7 0011&0100 == 0111

• A=A^0  (XOR by zeros has no effect)
• ~A = A ^ ~0  (XOR by 1's inverts all the bits)
• 0=A^A  (XOR by yourself creates 0's--used in cryptography)

The second property, that XOR by 1 inverts the value, is useful for flipping a set of bits.  Generally, XOR is used for equality testing (a^b!=0 means a!=b), controlled bitwise inversion, and crypto.

## Bitwise NOT: ~

Output bits are 1 if the corresponding input bit is zero.  E.g., ~011 == 111....111100.  (The number of leading ones depends on the size of the machine's "int".)

 As Ints As Bits ~0 == big value ~...0000 == ...1111

I don't use bitwise NOT very often, but it's handy for making an integer whose bits are all 1: ~0 is all-ones.

## Non-bitwise Logical Operators

Note that the logical operators &&, ||, and ! work exactly the same as the bitwise values, but for exactly one bit.  Internally, these operators map multi-bit values to a single bit by treating zero as a zero bit, and nonzero values as a one bit.  So
(2&&4) == 1 (because both 2 and 4 are nonzero)
(2&4) == 0 (because 2==0010 and 4 == 0100 don't have any overlapping one bits).