Floating Point Representation
Learn how computers store real numbers
Introduction
We have now learned to use two's complement which allows us to store negative numbers as well as positive numbers. However, this approach has limitations: it does't allow us to store real numbers (numbers with decimal parts) and makes storing large integers inefficient.
In this section, we will learn how a computer uses floating point representation to break numbers up into a mantissa and exponent.
Floating Point Representation
When working with floating point numbers each number is allocated a certain number of bits. Some of these might include:
- 16 bit
- 24 bit
- 32 bit
- 64 bit
This is the number of bits that will make up the entire number. This has to be split between the mantissa and the exponent.
We can do this in many different ways. For example we could split 32 bits between the mantissa and exponent as follows:
| Mantissa | Exponent |
|---|---|
| 24 bits | 8 bits |
| 16 bits | 16 bits |
| 8 bits | 24 bits |
The way in which we choose to split this will affect the range of numbers that we can store and the precision of those numbers as you will find out in the next section.
Mantissa
The mantissa is made up of 2 parts:
- Sign Bit
- Significand
The sign bit is a single bit that stores whether the number is positive or negative. If the number is positive the sign bit will be 0 and if the number is negative the sign bit will be 1.
The significand is the remainder of the mantissa and stores the significant digits of the number we are storing.
Exponent
The exponent stores the number of places the decimal point has been moved.
Examples
The 5 examples below which show you 5 different possible question you could be asked. Each differs slightly and in order to fully understand floating point representation you should ensure that you understand all 5 examples.
Example 1
Convert the following binary number into floating-point representation.
-11 0000 1111.1101 1
There are 16 bits for the mantissa (including the sign bit) and 8 bits for the exponent.
| Sign bit | Mantissa | Exponent |
|---|
Sign Bit
The sign bit is a single binary digit that stores whether the number is positive or negative.
If the number is positive, the sign bit is 0
If the number is negative, the sign bit is 1
In this case we can see that the number contains a negative sign and therefore the sign bit is 1.
| Sign bit | Mantissa | Exponent | 1 |
|---|
Mantissa
The mantissa stores the significant figures.
All floating point numbers are stored in the following format:
0·1...
For any floating point number, we must move the decimal point till it is left of the first 1
In our current example, the decimal point will move as follows:
This results in the decimal point being placed to the left of the first 1:
The numbers after the decimal point become the mantissa.
| Sign bit | Mantissa | Exponent | 1 | 1100 0011 1111 011 |
|---|
We have 16 bits available for the mantissa, but this includes the sign bit.
We have used 1 bit to store the sign bit which leaves 15 bits available for the rest of the mantissa.
The rest of the mantissa must be exactly 15 bits long, which it is!
Exponent
The exponent stores the number of places that the decimal point has been moved.
We can simply count this.
As we can see above, the decimal point has been moved 10 places to the left. Therefore, our exponent is 10!
We have 8 bits available for the exponent and will write 10 as an 8-bit binary number.
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
Therefore our mantissa is:
| Sign bit | Mantissa | Exponent | 1 | 1100 0011 1111 011 | 0000 1010 |
|---|
And that is us finished!
It is worth noting that the mantissa won't always be exactly 15 bits long and the exponent won't always be a positive number as we will see in the next 2 examples.
Example 2
Convert the following binary number into floating-point representation.
1111 0011.001
There are 16 bits for the mantissa (including the sign bit) and 8 bits for the exponent.
| Sign bit | Mantissa | Exponent |
|---|
Sign Bit
The sign bit is a single binary digit that stores whether the number is positive or negative.
Again, if the number is positive, the sign bit is 0. If the number is negative, the sign bit is 1
The number does not have a negative sign and therefore is positive. Our sign bit in this case is 0.
| Sign bit | Mantissa | Exponent | 0 |
|---|
Mantissa
The mantissa stores the significant figures.
All floating point numbers are stored in the following format:
0·1...
For any floating point number, we must move the decimal point till it is left of the first 1
In our current example, the decimal point will move as follows:
This results in the decimal point being placed to the left of the first 1:
The numbers after the decimal point become the mantissa.
| Sign bit | Mantissa | Exponent | 0 | 1111 0011 001 |
|---|
Again, we have 16 bits available for the mantissa, but this includes the sign bit.
We have used 1 bit to store the sign bit which leaves 15 bits available for the rest of the mantissa.
The rest of the mantissa must be exactly 15 bits long. Our current mantissa is only 11 bits long.
When this occurs, we pad the right side of the mantissa with 0's until it it the correct length.
In this case our mantissa becomes:
| Sign bit | Mantissa | Exponent | 0 | 1111 0011 0010 000 |
|---|
Exponent
Remember, the exponent stores the number of places that the decimal point has been moved.
As we can see above, the decimal point has been moved 8 places to the left. Therefore, our exponent is 8!
We have 8 bits available for the exponent and will write 8 as an 8-bit binary number.
Therefore our mantissa is:
| Sign bit | Mantissa | Exponent | 0 | 1111 0011 0010 000 | 0000 1000 |
|---|
And we are done!
Example 3
Convert the following binary number into floating-point representation.
-0.0001 1001 11
There are 16 bits for the mantissa (including the sign bit) and 8 bits for the exponent.
| Sign bit | Mantissa | Exponent |
|---|
Sign Bit
This time the number is negative and therefore our sign bit is 1.
| Sign bit | Mantissa | Exponent | 1 |
|---|
Mantissa
As stated before, all floating point numbers are stored in the following format:
0·1...
For any floating point number, we must move the decimal point till it is left of the first 1
In this example, the decimal point will move right instead of left
This results in the decimal point being placed to the left of the first 1:
The numbers after the decimal point become the mantissa.
| Sign bit | Mantissa | Exponent | 1 | 1100 111 |
|---|
Again, since we have 16 bits available for the mantissa and have used 1 bit for the sign bit, we have 15 bits to fill.
Currently we have only used 7 bits of the 15 available, so we will add 8 0's to the right.
| Sign bit | Mantissa | Exponent | 1 | 1100 1110 0000 000 |
|---|
Exponent
Remember, the exponent stores the number of places that the decimal point has been moved.
This time the decimal point has been moved 3 places to the right.
The direction the decimal point moved matters.
If the decimal point moves to the left, our exponent is positive.
If the decimal point moves to the right, our exponent is negative.
This time the decimal point has been moved 3 places to the right, so our exponent is -3.
There are 8 bits available for the exponent. Therefore we will write -3 as an 8-bit two's compliment number.
| Sign bit | Mantissa | Exponent | 1 | 1100 1110 0000 000 | 1111 1101 |
|---|
Finished.
Example 4
Convert the following binary number into floating-point representation.
1100 1111 1111. 1101 11
There are 16 bits for the mantissa (including the sign bit) and 8 bits for the exponent.
| Sign bit | Mantissa | Exponent |
|---|
Sign Bit
This time the number is positive and therefore our sign bit is 0.
| Sign bit | Mantissa | Exponent | 0 |
|---|
Mantissa
Let's move the decimal point and find the mantissa.
This makes our mantissa:
| Sign bit | Mantissa | Exponent | 0 | 1100 1111 1111 1101 11 |
|---|
However, there are only 16 bits available for the mantissa. 1 is used for the sign bit leaving us with 15 for the rest of the mantissa.
We only have 15 bits available but have used 18 bits! This is an issue, we can't use more bits than are available.
In this case the mantissa cuts off when it hits the limit of the number of bits available.
The mantissa used above is 3 bits over the allocation, therefore the last 3 bits are cut off.
| Sign bit | Mantissa | Exponent | 0 | 1100 1111 1111 110 |
|---|
This leads to a loss of data and is one of the issues that can arise when there are not enough bits allocated to the mantissa.
Exponent
The exponent stores the number of places that the decimal point has been moved.
This time the decimal point has been moved 12 places to the left.
We are moving left so our number will be positive.
As stated in the question, the exponent is allocated 8 bits, therefore we will write 12 as an 8 bit binary number.
| Sign bit | Mantissa | Exponent | 0 | 1100 1111 1111 110 | 0000 1100 |
|---|
Done.
Example 5
Convert the following binary number into floating-point representation.
-0.0011 1101 1100 1111
There are 28 bits for the mantissa (including the sign bit) and 4 bits for the exponent.
| Sign bit | Mantissa | Exponent |
|---|
Sign Bit
This time the number is negative and therefore our sign bit is 1.
| Sign bit | Mantissa | Exponent | 1 |
|---|
Mantissa
This time our mantissa has been allocated 28 bits in total. 1 bit will be used for the sign bit leaving 27 bits for the rest of the mantissa.
Again we will move the decimal point to achieve the following format:
0·1...
The numbers after the decimal point become the mantissa.
| Sign bit | Mantissa | Exponent | 1 | 1111 0111 0011 11 |
|---|
There are 23 bits allocated so we will fill the remaining space with 0's.
| Sign bit | Mantissa | Exponent | 1 | 1111 0111 0011 1100 0000 0000 000 |
|---|
Exponent
The exponent stores the number of places that the decimal point has been moved.
This time the decimal point has been moved 2 places to the right.
Since we are moving to the right, our exponent will be negative.
We have been allocated 4 bits for our exponent. Therefore we will write -2 as a 4-bit two's compliment number.
| Sign bit | Mantissa | Exponent | 1 | 1111 0111 0011 1100 0000 0000 000 | 1110 |
|---|
We have finished the final example.