# CS140 Lecture notes -- Bit Arithmetic

• Jim Plank
• Directory: ~cs140/www-home/notes/Bits
• Lecture notes: http://www.cs.utk.edu/~cs140/notes/Bits
• Thu Nov 5 13:08:14 EST 1998
Note, I am not going to go over things in this level of detail in class, so if class left you a bit fuzzy on bit arithmetic, go over these notes thoroughly.

## Bit Arithmetic

C has operations for bit arithmetic about which students are often unaware. As you know, each data type is a certain number of bytes, and each byte is eight bits. You can kind of see the bit structure of a byte by printing it in hexidecimal or in octal. To truly see the bits, you have write some code.

Look at the program printbits.c:

```
print_bits(unsigned char c)
{
....
}

main()
{
unsigned char c;
int i;

while (1) {
printf("Enter an integer between 0 and 255: ");
fflush(stdout);
if (scanf("%d", &i) != 1) exit(0);
if (i < 0 || i > 255) {
} else {
c = (unsigned char) i;
print_bits(c);
putchar('\n');
}
}
}
```
This repeatedly takes an integer between 0 and 255, puts it into an unsigned char (i.e. one byte), and then calls print_bits() on it. Print_bits() prints out the bits. Don't worry about how it is implemented yet. The output should be straightforward -- the bits look as you think they should:
```UNIX> printbits
Enter an integer between 0 and 255: 0
00000000
Enter an integer between 0 and 255: 1
00000001
Enter an integer between 0 and 255: 2
00000010
Enter an integer between 0 and 255: 3
00000011
Enter an integer between 0 and 255: 4
00000100
Enter an integer between 0 and 255: 5
00000101
Enter an integer between 0 and 255: 255
11111111
Enter an integer between 0 and 255: 254
11111110
Enter an integer between 0 and 255: < CNTL-D >
UNIX>
```

### Bitwise-and

The bit operations in C operate on these bits. The first operation is bitwise-and. It is the ampersand. The bitwise-and of two numbers simply goes through the bit position and sets that position to 1 if the two bits in that position are 1, and 0 otherwise. Look at and.c. This takes two bytes and prints out their bitwise-and. You'll note, 0 bitwise-and anything is zero. 255 bitwise-and x is x, because all the bits of 255 are one.
```UNIX> and
Enter two integers between 0 and 255: 1 0
00000001 & 00000000 = 00000000
Enter two integers between 0 and 255: 1 1
00000001 & 00000001 = 00000001
Enter two integers between 0 and 255: 1 2
00000001 & 00000010 = 00000000
Enter two integers between 0 and 255: 1 3
00000001 & 00000011 = 00000001
Enter two integers between 0 and 255: 2 3
00000010 & 00000011 = 00000010
Enter two integers between 0 and 255: 0 40
00000000 & 00101000 = 00000000
Enter two integers between 0 and 255: 255 40
11111111 & 00101000 = 00101000
Enter two integers between 0 and 255: 0 89
00000000 & 01011001 = 00000000
Enter two integers between 0 and 255: 255 89
11111111 & 01011001 = 01011001
Enter two integers between 0 and 255: < CNTL-D >
UNIX>
```
If a number is all zeros except for a one in the i-th position (this means that the number is 2^(i-1)) then that number bitwise-and x is equal to zero or that number, depending on whether x has a bit set in the i-th position. Therefore (4 & 7) equals 4 since 7 has its bit in the third position set.

Similarly, if you are interested in what the lowest i bits of a byte are, you can take that byte bitwise-and (2^i)-1. For example, if you want to see what the lowest three bits are of a byte, you take that byte bitwise-and (2^3)-1 = 7. For example, the lowest three bits of 89 are (89&7)=1:

```UNIX> and
Enter two integers between 0 and 255: 4 7
00000100 & 00000111 = 00000100
Enter two integers between 0 and 255: 4 2
00000100 & 00000010 = 00000000
Enter two integers between 0 and 255: 89 7
01011001 & 00000111 = 00000001
Enter two integers between 0 and 255: 91 7
01011011 & 00000111 = 00000011
Enter two integers between 0 and 255: < CNTL-D >
UNIX>
```

### Bitwise-or

Bitwise-or works like bitwise-and, only it sets the bit to one if either of the numbers has its bit in that position set to one. It's operator is the pipe character (|). You'll note that (0|x) = x, and (255|x) = 255.
```UNIX> or
Enter two integers between 0 and 255: 0 1
00000000 | 00000001 = 00000001
Enter two integers between 0 and 255: 1 2
00000001 | 00000010 = 00000011
Enter two integers between 0 and 255: 1 3
00000001 | 00000011 = 00000011
Enter two integers between 0 and 255: 0 40
00000000 | 00101000 = 00101000
Enter two integers between 0 and 255: 255 40
11111111 | 00101000 = 11111111
Enter two integers between 0 and 255: 0 89
00000000 | 01011001 = 01011001
Enter two integers between 0 and 255: 255 89
11111111 | 01011001 = 11111111
Enter two integers between 0 and 255: < CNTL-D >
UNIX>
```

### Other bitwise operations

There is also bitwise exclusive-or, which returns whether the two bits are different. Its operator is the carat (^). And there is bitwise complement, which is unary, and changes zeros to ones and ones to zero:
```UNIX> exor
Enter two integers between 0 and 255: 1 2
00000001 ^ 00000010 = 00000011
Enter two integers between 0 and 255: 1 3
00000001 ^ 00000011 = 00000010
Enter two integers between 0 and 255: 255 0
11111111 ^ 00000000 = 11111111
Enter two integers between 0 and 255: 255 255
11111111 ^ 11111111 = 00000000
Enter two integers between 0 and 255: 1 1
00000001 ^ 00000001 = 00000000
Enter two integers between 0 and 255: < CNTL-D >
UNIX> complement
Enter an integer between 0 and 255: 0
~ 00000000 = 11111111
Enter an integer between 0 and 255: 1
~ 00000001 = 11111110
Enter an integer between 0 and 255: 40
~ 00101000 = 11010111
Enter an integer between 0 and 255: 255
~ 11111111 = 00000000
Enter an integer between 0 and 255: < CNTL-D >
UNIX>
```

### Bit shifting

Finally, there are the bit shift operators: left shift (<<) and right shift (>>). They shift the bits a specified number of positions left and right respectively:
```UNIX> leftshift
Enter an integer and a shift amount: 1 0
00000001 <<  0 = 00000001
Enter an integer and a shift amount: 1 1
00000001 <<  1 = 00000010
Enter an integer and a shift amount: 1 2
00000001 <<  2 = 00000100
Enter an integer and a shift amount: 1 3
00000001 <<  3 = 00001000
Enter an integer and a shift amount: 1 4
00000001 <<  4 = 00010000
Enter an integer and a shift amount: 3 1
00000011 <<  1 = 00000110
Enter an integer and a shift amount: 3 0
00000011 <<  0 = 00000011
Enter an integer and a shift amount: 0 1
00000000 <<  1 = 00000000
Enter an integer and a shift amount: 3 8
00000011 <<  8 = 00000000
Enter an integer and a shift amount: < CNTL-D >
UNIX> rightshift
Enter an integer and a shift amount: 1 0
00000001 >>  0 = 00000001
Enter an integer and a shift amount: 1 1
00000001 >>  1 = 00000000
Enter an integer and a shift amount: 8 1
00001000 >>  1 = 00000100
Enter an integer and a shift amount: 8 2
00001000 >>  2 = 00000010
Enter an integer and a shift amount: 8 3
00001000 >>  3 = 00000001
Enter an integer and a shift amount: 255 1
11111111 >>  1 = 01111111
Enter an integer and a shift amount: 255 7
11111111 >>  7 = 00000001
Enter an integer and a shift amount: < CNTL-D >
UNIX>
```
You'll note, left shifting by 1 multiplies by 2, and right shifting by one divides by two. In general, left shifting by i multiplies by 2^i, and right shifting by i divides by 2^i.

However, a word of caution. Yes, left-shifting by one multiplies by 2 faster than doing x*2. However, do not do this if you want to multiply by two unless your code really has to be optimally fast, or unless it's pretty clear what you're doing. Otherwise, your code becomes extremely hard to read.

Now that you know all this, implementing print_bits() is pretty simple:

```print_bits(unsigned char c)
{
unsigned int i;

for (i = 1 << 7; i != 0; i >>= 1) {
if (i&c) putchar('1'); else putchar('0');
}
}
```
Yes, I could have done:
```  for (i = 1 << 7; i != 0; i >>= 1) {
putchar('0'+((i&c)>0));
}
```