Signed Binary Numbers
So far, we've treated all binary numbers in this discussion as unsigned numeric values. However, we can also use signed binary numbers.
What makes one number signed, and another number unsigned? Whether a number is a signed number or an unsigned number depends solely on how we treat the number in our program. If we perform mathematics and check the SIGN Flag in the Flag Register or test the MSB for a 0 or 1, we are treating the value as a signed number.
The binary number 0000 0000 represents zero, 0000 0001 represents one, 0000 0010 represents two, and so on toward the largest number that we care to represent.
What about negative numbers? How can we place a minus sign in a binary position? Well, we can't. So we assign a bit, the MSB, as a SIGN bit.
The rule for signed and unsigned binary numbers is simple:
The number of numeric values that can be represented by any
The table below contains the range of values for signed and unsigned binary numbers organized on a Byte, 16-bit Word, and 32-bit Double Word boundary.
Size | Byte | Word | Double Word |
Number of Values | 256 | 65,536 | 4,294,967,296 |
Unsigned Number | 0 to 255 | 0 to 65,535 | 0 to 4,294,967,295 |
Signed Number | -128 to 127 | -32,768 to 32,767 | -2,147,483,648 to 2,147,483,647 |
For the Byte, we represent the negative values in the range -128 through -1 and the positive values in the range 0 through 127 with a single eight bit byte.
With a 16-bit word we can represent negative values in the range -32,768 through -1, and positive values in the range 0 through +32,767.
With a 32-bit double word we can represent negative values in the range
-2,147,483,648 through -1 and the positive values in the range 0 through
+2,147,483,647.
Negative Binary
Numbers - The 2's Complement
We do not just place a 1 in the MSB of a binary number to make it negative. We must take the 2's Complement of the number. Taking the 2's Complement of the number will cause the MSB to become 1.
To obtain the 2's complement of a number is a two step process.
The complement of a binary number, also known as the 1's complement, requires us to change every logic 1 bit in a number to a logic 0, and every logic 0 bit to a logic 1. Let's find the 1's complement of 36H or 0011 0110 in binary. In the following table, the 1's Complement is shown in RED.
Number Format | D7 | D6 | D5 | D4 | D3 | D2 | D1 | D0 | ||
Unsigned Number | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | ||
1's Complement | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
To obtain the 2's complement of a binary number, we must first obtain the 1's complement of the number and then add 1. In the following table, the 1's complement is RED and the 2's Complement is in BLUE. The 2's complement of 36H or 0011 0110 in binary is:
Number Format | D7 | D6 | D5 | D4 | D3 | D2 | D1 | D0 | ||
Unsigned Number | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | ||
1's Complement | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | ||
2's Complement | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
If we are using signed binary numbers and the MSB is already logic 1, it means the value is the 2's complement of the number. The actual numeric value can be determined by taking the 2's complement of the number and keeping track of the minus sign mentally.
Find the decimal value of the signed binary number 0FFH.
Number Format | D7 | D6 | D5 | D4 | D3 | D2 | D1 | D0 | ||
Signed Number | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||
1's Complement | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ||
2's Complement | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
Therefore, 0FF is a signed number that has the value in decimal of -1.
Subtraction of Binary Numbers -
Using 2's Complement Addition
If we add the 2's Complement of a signed number to another signed number, we are performing the mathematical operation of subtraction. This is, in fact. how many 8-bit microprocessors actually perform subtraction, they perform 2's complement addition.
Let us see if 0FFH is really -1. If we add it to 01H, we expect the result to be 0. One special condition about twos complement addition, overflows (a CARRY OUT which SETS the CARRY FLAG) from the register are ignored. So let's see what happens.
In Decimal | CO | D7 | D6 | D5 | D4 | D3 | D2 | D1 | D0 | ||
1 | X | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | ||
-1 | X | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
The last row of the table shows that the result is 0. The 1 in the CO (CARRY OUT) is ignored. If we ignore the CO, the result is 0000 0000.
Okay, so it works for a byte. How about a Word? It is the same procedure. Only the MSB is used as the sign bit, so the MSB of the low order byte is a weighted position bit. For 16-bit numbers: