Numeric Data Representation in Modern Computer System

Integer, fix-point number, floating-point number

This is a summary of MIT6.5940, lecture 5, data types part.

1. Integer

1.1 Unsigned Integer

For an N-bit unsigned integer, its range is $[0, 2^N - 1]$

1.2 Signed Integer

1.2.1 Sign-Magnitude Representation

For N-bit, the most left(highest) bit is a sign bit. Its range is $[-(2^{n-1}-1), 2^{n-1}-1]$

There are two ways to represent 0

• 10000000
• 00000000

1.2.2 Two’s Complement Representation

sign bit

• Most significant bit (MSB) is the sign bit.
• if MSB is 0, the number is positive or zero.
• if MSB is 1, the number is negative.

range(for N-bit)

• The min is where MSB is 1 and the rest is 0, which is $-2^{n-1}$.
• The max is where MSB is 0 and the rest is 1, which is $2^{n-1}-1$.

Two’s complement has only one representation for zero: 00000000.

Steps to obtain Two’s complement:

Positive Numbers:

• Represent the number in binary as usual.

Negative Numbers:

• Start with the binary representation of the absolute value.
• Invert all the bits (change 0s to 1s and 1s to 0s).
• Add 1 to the least significant bit (LSB).

For example:

Positive Number (e.g., +5):

1. Convert +5 to binary: 0101
2. The result is 0101, which is the binary representation in two’s complement.

Negative Number (e.g., -5):

1. Convert 5 to binary: 0101
2. Invert the bits: 1010
3. Add 1: 1010 + 1 = 1011
4. The result is 1011, which is the binary representation of -5 in two’s complement.

Two’s complement makes the arithmetic simpler and hardware implementation friendly.

Addition of 4-bit Two's complement numbers

3+(-5)

3 is "0011"
5 is "1011"

  0011
+ 1011
------
  1110

"1110" is -2 in Two's complement representation.

2. Fixed-Point Number

source: https://www.dropbox.com/scl/fi/eos92o2fgys6gk0gizogl/lec05.pdf?rlkey=2hohvi8jcvjw3f8m8vugfa2mz&e=1&dl=0

Example with negative numbers.

-5.5

In 8 bit representation, one possible representation could be 0101.1000
Invert bits: 1010.0111
Add 1 to the LSB: 1010.0111 + 0.0001 = 1010.1000

1010.1000 -> 1x(-2^(3)) + 1x2^(1) + 1x2^(-1) = -8 + 2 + 0.5 = -5.5

3. Floating-Point Number

3.1 32-bit floating point number

The number it can represent:
$$
(-1)^{sign} \times (1 + mantissa) \times 2^{exponent-127}
$$

Q: How can we represent zero in 32-bit floating point number?

The equation above only applies to cases where exponent is not all zeros. If exponent is all zeros, then it follows another equation.
$$
(-1)^{sign} \times mantissa \times 2^{1-127}
$$
So, we have two ways to represent zero, where we let mantissa part to be 0.

3.2 Half Precision (FP16)

$$
(-1)^{sign} \times (1 + mantissa) \times 2^{exponent-15}
$$

4. INT4

Negative numbers are represented using Two’s complement.

Binary | Decimal

0000 -> 0
0001 -> 1
0010 -> 2
0011 -> 3
0100 -> 4
0101 -> 5
0110 -> 6
0111 -> 7
1000 -> -8
1001 -> -7
1010 -> -6
1011 -> -5
1100 -> -4
1101 -> -3
1110 -> -2
1111 -> -1

5. FP4

References

• https://www.dropbox.com/scl/fi/eos92o2fgys6gk0gizogl/lec05.pdf?rlkey=2hohvi8jcvjw3f8m8vugfa2mz&e=1&dl=0