Basic Arithmetic (adding and subtracting)
Digital logic to show add/subtractBoolean algebra abstraction of physical, analog circuit behavior
1
0
CPUcomponents – ALUlogic circuitslogic gatestransistors
Digital Logic
A B A * B0 0 00 1 01 0 01 1 1
A B A B0 0 00 1 01 0 01 1 1
AB
out
outAB
and (* or ^)
or ( )
Digital Logic
A out
outAB
not (~, ⌐, −)
xor ( )
A ~A0 11 0
A BA B
0 0 00 1 11 0 11 1 0
Digital Logic
AB
out
outAB
nand
nor
A B0 0 10 1 11 0 11 1 0
B*A
A BA nor
B0 0 10 1 01 0 01 1 0
Digital Logic
Given the truth table:
F = + + +
Sum of products from the truth table. Often we can simplify.
A B F
0 0 10 1 11 0 11 1 1
BABABABA
BA BA BA BA
Unsigned ArithmeticBinary Addition
A B cout sum
0 0 0 10 1 0 11 0 0 11 1 1 0
Input Output
A B S C
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
cout
sumAB
Half Adder
Unsigned Arithmetic
The half adder is an example of a simple digital circuit built from two logic gates.
half adder logic - two one-bit inputs (a, b) and two one-bit outputs (carry_out, sum) a 0 0 1 1 + b + 0 + 1 + 0 +1
carry_out sum 00 01 01 10
carry_out = a and b sum = a xor b
Unsigned Arithmetic
The problem with a half-adder is that it doesn't handle carries.
Consider adding the following two numbers: 1 1 1 + 0 1 1
When we add the two numbers, we get1 1
1 1 1 + 0 1 1 (1) 0 1 0
Look at the middle and leftmost columns. You add 3 bits. Half adders can only add two bits.
Unsigned Arithmetic
For adding multi-bit fields/words, e.g., 4 bits a_3 a_2 a_1 a_0 + b_3 b_2 b_1 b_0 --------------------------------------- sum_3 sum_2 sum_1 sum_0
we also need to add a carry_in with a_i and b_i, where i>0
Unsigned Arithmetic
A full adder for a_i + b_i + carry_in is given in the figure below.– three one-bit inputs (a, b, carry_in) and– two one-bit outputs (carry_out, sum)– cascade two half adders (sum output bit of first
attached to one input line of the other) and then or together the carry_outs
InputOutpu
tC A B S C0 0 0 0 01 0 0 1 00 1 0 1 01 1 0 0 10 0 1 1 01 0 1 0 10 1 1 0 11 1 1 1 1
Cin
AB
sum
Cout
Full Adder• An n-bit adder built by connecting n full adders• carries propagate from right to left (i.e., connect
the carry_out of an adder to the carry_in of the adder in the next leftmost bit position
• the initial, that is, rightmost, carry_in is zero)
overflow occurs when a number is too large to represent for unsigned arithmetic, overflow occurs when a carry out occurs from the most significant (i.e., leftmost) bit position
Full Adder
(there are faster forms of addition hardware where the carries do not have to propagate from one side to the other, e.g., carry-lookahead adder)
Signed Arithmetic
fundamental idea #1 finite width arithmetic- modulus rn, where r is radix, n is number of digits wide
- wraps around from biggest number to zero, ignoring overflow
e.g., 4-bit arithmetic => modulus is 24 = 16 0, 1, 2, ..., 15 then wrap around back to 0
thus an addition of rn to an n-digit finite width value has no effect on the n-digit value
overflow occurs when a number is too large to represent for unsigned arithmetic, overflow occurs when a carry out occurs from the most significant (i.e., leftmost) bit position (there are faster forms of addition hardware where the carries do not have to propagate from one side to the other, e.g., carry-lookahead adder)
Signed Arithmetic
fundamental idea #2 subtraction is equivalent to adding the negative of number
e.g., a - b = a + (-b)
observation a - b == a - b + rn == a + (rn - b) \______/ \_______/ #1 #2 \______/ this term is our representation for (-b)
it turns out that we can more easily perform rn - b than a - b
Signed Arithmetic
digit complement for n digits == (rn - 1) - number in binary, this is called one's complement and equals a value of n ones minus the bits of the number
for binary, one's complement (2n - 1 - number) is equivalent to inverting each bit
in decimal, this is called nine's complement and equals a value of n nines minus the digits of the number
in hexadecimal, this is n f's (fifteens) minus the digits of the number
Signed Arithmetic
radix complement for n digits == (rn - 1) - number + 1
two's complement in binary ten's complement in decimal
for binary, two's complement (2n - 1 - number + 1) is equivalent to inverting each bit and adding one
One's complement format of a number
– Change the number to binary, ignoring the sign.– Add 0s to the left of the binary number to make a
total of n bits– If the sign is positive, no more action is needed.– If the sign is negative, complement every bit (i.e.
change from 0 to 1 or from 1 to 0)
Signed Arithmetic
Most computers today use 2's complement representation for negative numbers.
The 2's complement of a negative number is obtained by adding 1 to the 1's complement. i.e.
Two's complement format of a number– Change the number to binary, ignoring the sign.– Add 0s to the left of the binary number to make a
total of n bits– If the sign is positive, no more action is needed.– If the sign is negative, complement every bit (i.e.
change from 0 to 1 or from 1 to 0) and add 1
Signed Arithmetic
Examples converting to two’s complement representation:For -13, 8-bit representation
00001101 base integer11110010 1's complement +111110011 2's complement
For -227, 12-bit representation
000011100011 base integer111100011100 1's complement +1111100011101 2's complement
Signed Arithmetic
Converting from two's complement to decimal:
• If the sign bit is 0, convert the binary number to decimal.
• If the sign bit is 1– complement each bit– add 1– convert the binary number to decimal– put a minus sign in front
Signed Arithmetictwo's complement encoding - note one zero and asymmetric range first (leftmost) bit is sign bit (1 => -)
binary sign magnitude one's complement two's complement
0000 +0 +0 0 0001 +1 +1 +1 0010 +2 +2 +2 0011 +3 +3 +3 0100 +4 +4 +4 0101 +5 +5 +5 0110 +6 +6 +6 0111 +7 +7 +7 1000 -0 -7 -8 1001 -1 -6 -7 1010 -2 -5 -6 1011 -3 -4 -5 1100 -4 -3 -4 1101 -5 -2 -3 1110 -6 -1 -2 1111 -7 -0 -1
Signed ArithmeticWe can easily make a full adder do subtraction by adding an inverter in front of each b sub i and setting carry into the rightmost adder to one
Signed Arithmeticrange for n-bit field: unsigned is [ 0, 2n - 1 ]2's compl. signed is [ -2n-1, 2n-1 - 1 ]
signed overflow occurs whenever the sign bits of the two operands agree, but the sign bit of the result differs (i.e., add two positives and result appears negative, or add two negatives and result appears nonnegative)range diagrams for three bitsunsigned
signed(2's compl)
000 001 010 011 100 101 110 111
|------- |------- |------- |------- |------- |------- |-------| 0 1 2 3 4 5 6
7
100 101 110 111 000 001 010 011
|------- |------- |------- |------- |------- |------- |-------|-4 -3 -2 -1 0 +1 +2
+3
b2 b1 b0
sign
b1
b0
Signed Arithmeticmodulo arithmetic (keep adding +1 and wrap around)
000 001 010 011 100 101 110 111
(unsigned) 0 1 2 3
4 5 6 7
(or 2's compl)
0
+1
+2
+3
-4
-3
-2
-1
^^--carry occurs on wrap around
Signed Arithmetic3-bit examples
bits unsigned signed 111 = 7 = (–1)
+001 = +1 = +(+1) ----- --- -------000 0 (0)(carry) OVF ^^^-- this is what the ALU computes for either unsigned or signed. but, while it is an unsigned overflow, it is CORRECT for signed
Signed Arithmetic3-bit examples
Example 2 bits unsigned signed 011 3 = (+3) +001 = +1 = +(+1) ----- -- ----- 100 4 (–4) OVF
^^^-- this is what the ALU computes for either unsigned or signed, but, while it is correct for unsigned, it is SIGNED OVERFLOW!
Signed Arithmetic
16-bit signed (2's complement) examples
in 16-bit arithmetic, we can represent values as four hex digits;if the leading hex digit is between 0 and 7 (thus it has a leading bit of 0), it is a nonnegative value; if the leading hex digit is between 8 and f (thus it has a leading bit of 1), it is a negative value
signed overflow occurs if a. (+) added with (+) gives a (-), or b. (-) added with (-) gives a (+)
Signed Arithmetic
hexadecimal hexadecimal decimal
0x7654 = 0x7654 = (+30292)
+0xffed = +(-0x13) = +( –19)
0x7641 0x7641 (+30273) (carry) carry occurs but there is no signed overflow (thus carry is ignored)
(+) added with (-) cancels out, so signed overflow is not possible
Signed Arithmetic
hex decimal 0x7654 = (+30292)+0x1abc = +( + 6844) ---------- ------------ 0x9110 (-28400) should be 37136, but is >
max positive 32767
no carry occurs but there is signed overflow (+) added with (+) giving (-) => SIGNED OVERFLOW!
Signed Arithmetic
hex decimal 0x7654 = (+30292)+0x1abc = +( + 6844)
0x9110 (-28400) should be 37136, but is > max positive 32767
no carry occurs but there is signed overflow (+) added with (+) giving (-) => SIGNED OVERFLOW!
Signed Arithmetic
hexadecimal 0x7654 change subtraction to addition by ffff -0xff8d taking two's complement of 0xff8d -ff8d
0072+ 1
0073
hexadecimal hexadecimal decimal
0x7654 = 0x7654 = (+30292)
-0xff8d = +0x0073 = +( +115) 0x76c7 = (+30407)
no carry occurs and no signed overflow (+) added with (+) giving (+) => no signed overflow