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One/Two’s complement/Hex
One’s
Binary
Decimal
0111
0110
0101
0100
0011
0010
0001
0000
1111
1110
1101
1100
1011
1010
1001
1000
+7
+6
+5
+4
+3
+2
+1
+0
-0
-1
-2
-3
-4
-5
-6
-7
Here is a quick summary of how to find the 1's complement representation of any decimal
number x.
1. If x is positive, simply convert x to binary.
2. If x is negative, write the positive value of x in binary
3. Reverse each bit.
Example –convert 5 to one’s complement and vice versa
5
0101
-5
0101
1010 – reverses each bit
-5
1010
Reverses to
0101 which is 5
1
As an example, the ones' complement form of 00101011 (43) becomes 11010100 (−43). The
range of signed numbers using ones' complement is represented by −(2N−1−1) to (2N−1−1) and
+/−0. A conventional eight-bit byte is −12710 to +12710 with zero being either 00000000 (+0) or
11111111 (−0).
Binary Addition
To add two numbers represented in this system, one does a conventional binary addition, but it is
then necessary to add any resulting carry back into the resulting sum. To see why this is
necessary, consider the following example showing the case of the addition of −1 (11111110) to
+2 (00000010).
binary decimal
11111110 -1
+ 00000010 +2
............ ...
1 00000000 0 <-- not the correct answer
1 +1 <-- add carry
............ ...
00000001
1 <-- correct answer
VERY IMPORTANT(EVERYTHING STEMS FROM CALC BELOW FOR ONE’S vs.TWO’S COMPLEMENT!!!
Adding binary numbers is a very simple task, and very similar to the longhand addition of
decimal numbers. As with decimal numbers, you start by adding the bits (digits) one column, or
place weight, at a time, from right to left. Unlike decimal addition, there is little to memorize in
the way of rules for the addition of binary bits:
0
1
0
1
1
+
+
+
+
+
0
0
1
1
1
=
=
=
=
+
0
1
1
10
1 = 11
Just as with decimal addition, when the sum in one column is a two-bit (two-digit) number, the
least significant figure is written as part of the total sum and the most significant figure is
"carried" to the next left column. Consider the following examples:
.
.
.
.
.
1001101
+ 0010010
--------1011111
11 1 <--- Carry bits -----> 11
1001001
1000111
+ 0011001
+ 0010110
----------------1100010
1011101
2
Two’s Complement
The two's complement of a binary number is defined as the value obtained by subtracting the
number from a large power of two (specifically, from 2N for an N-bit two's complement). The
two's complement of the number then behaves like the negative of the original number in most
arithmetic, and it can coexist with positive numbers in a natural way.
A two's-complement system or two's-complement arithmetic is a system in which negative
numbers are represented by the two's complement of the absolute value; this system is the most
common method of representing signed integers on computers. In such a system, a number is
negated (converted from positive to negative or vice versa) by computing its two's complement.
An N-bit two's-complement numeral system can represent every integer in the range −2N−1 to
+2N−1−1.
Two's complement
0100
0011
0010
0001
0000
1111
1110
1101
1100
1111 1111
− 0101 1111
===========
1010 0000
+
1
===========
1010 0001
Decimal
4
3
2
1
0
−1
−2
−3
−4
255
− 95
=====
(ones' complement)
160
+
1
=====
(two's complement)
161
Example: Convert 4 to two’s complement value
4
0100
reverse (one’s complement)
1011
+ 1 (perform binary add.)
=====
Which is
1011
+0001
=====
1100
3
1100 -4
0100 4
Example -4
1100
0011 (reverse)
3
+
1
1 (add 1)
-----0100
4
HEX
Hexadecimal is base 16.
Base 16 is where the 'numbers' you can use are zero through to the letter F (0123456789ABCDEF). i.e.
the decimal value for '1' is represented in hexadecimal as '1' but the hexadecimal value of '15' (decimal) is
shown as 'F' (hexadecimal) and the value of '17' (decimal) is '11' in Hexadecimal.
Decimal
Hex
Decimal
Hex
Decimal
Hex
1
1
11
B
30
1E
2
2
12
C
40
28
3
3
13
D
50
32
4
4
14
E
60
3C
5
5
15
F
70
46
6
6
16
10
80
50
7
7
17
11
90
5A
8
8
18
12
100
64
9
9
19
13
500
1F4
10
A
20
14
1000
3E8
4
Hex to Decimal
Steps:
1.
2.
3.
4.
5.
6.
7.
Get the last digit of the hex number, call this digit the currentDigit.
Make a variable, let's call it power. Set the value to 0.
Multiply the current digit with (16^power), store the result.
Increment power by 1.
Set the the currentDigit to the previous digit of the hex number.
Repeat from step 3 until all digits have been multiplied.
Sum the result of step 3 to get the answer number.
Example 1
Convert the number 1128 HEXADECIMAL to DECIMAL
MULTIPLICATION RESULT NOTES
Start from the last digit of the number. In this
case, the number is 1128. The last digit of that
8 x (16^0)
8
number is 8. Note that the power of 0 of any
number is always 1
Process the previous, which is 2. Multiply that
2 x (16^1)
32
number with an increasing power of 16.
Process the previous digit, which is 1, note that
1 x (16^2)
256
16^2 means 16 x 16
Process the previous digit, which is 1, note that
1 x (16^3)
4096
16^3 means 16 x 16 x 16
Here, we stop because there's no more digit to
process
This number comes from the sum of the
ANSWER
4392
RESULTS
(8+32+256+4096)=4392
Once discerned, notice that the above process is essentially performing this
calculation:
1x(16^3) + 1x(16^2) + 2x(16^1) + 8x(16^0)
5
When doing this by hand, it is easier to start backward is because:


Counting the number of digits takes extra time, and you might count
wrongly.
If you don't remember what a particular value of a power-of-16 is, it's
easier to calculate it from the previous power value. For instance, if you
don't remember what the value of 16^3 is, then just multiply the value of
16^2 (which you'll likely already have if you started backward) with 16.
Example 2
Convert the number 589 HEXADECIMAL to DECIMAL
MULTIPLICATION
9 x (16^0)
8 x (16^1)
5 x (16^2)
RESULT
9
128
1280
ANSWER
1417
If you want to be a speed counter, it's beneficial to memorize the values of the
smaller power of 16s, such as in this table
POWER OF 16s
16^0
16^1 = 16
16^2 = 16x16
16^3 = 16x16x16
16^4 = 16x16x16x16
RESULT
1
16
256
4096
65536
Example 3
Convert the number 1531 HEXADECIMAL to DECIMAL
(This time, let's use the table of the power-of-16s above.)
MULTIPLICATION
1x1
3 x 16
5 x 256
1 x 4096
RESULT
1
48
1280
4096
6
ANSWER
5425
7
Decimal to Hex
Steps:
1. Divide the decimal number by 16. Treat the division as an integer
division.
2. Write down the remainder (in hexadecimal).
3. Divide the result again by 16. Treat the division as an integer
division.
4. Repeat step 2 and 3 until result is 0.
5. The hex value is the digit sequence of the remainders from the last to
first.
Note: a remainder in this topic refers to the left over value after
performing an integer division.
HEXADECIMAL 0
1
2
3
4
5
6
7
8
9
A
DECIMAL
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
0
B
C D E
F
Example 1
Convert the number 1128 DECIMAL to HEXADECIMAL
NOTES
DIVISION RESULT
REMAINDER (in
HEXADECIMAL)
1128 / 16 70
8
Start by dividing the
number by 16.
In this case, 1128
divided by 16 is
70.5. So the integer
division result is 70
(throw out anything
after the decimal
point).
The remainder is
(70.5 - 70) multiplied
with 16; or (0.5 times
16), which is 8.
Then, divide the result 70 / 16
again by 16
4
6
8
(the number 70 on the
DIVISION
column comes from
the previous
RESULT).
In this case,
70/16=4.375. So
the integer division
result is 4 (throw out
anything after the
decimal point)
The remainder is
(0.375 multiplied
with 16, which is 6.
Repeat. Note here
that 4/16=0.25. So
the integer division
result is 0.
4 / 16
0
4
The remainder is
(0.25-0) multiplied
with 16, which is 4.
Stop because the
result is already 0 (0
divided by 16 will
always be 0)
Well, here is the
answer. These
numbers come from
the REMAINDER
column values (read
from bottom to top)
468
Side note: You can get the remainder of a division using the
Modulus or % operator. Ie: 1128%16=8.
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Example 2
Convert the number 256 DECIMAL to HEXADECIMAL
DIVISION
RESULT
REMAINDER (in HEX)
256 / 16
16
0
16 / 16
1
0
1 / 16
0
1
ANSWER
100
Example 3
Convert the number 921 DECIMAL to HEXADECIMAL
DIVISION
RESULT
REMAINDER (in HEX)
921 / 16
57
9
57 / 16
3
9
3 / 16
0
3
ANSWER
399
10
Example 4
Convert the number 188 DECIMAL to HEXADECIMAL
DIVISION
RESULT
REMAINDER
(in HEX)
188 / 16
11
C (12 decimal)
11 / 16
0
B (11 decimal)
ANSWER
BC
Note that here, the answer would not be 1112, but BC. Remember to
write down the remainder in hex, not decimal.
Example 5
Convert the number 590 DECIMAL to HEXADECIMAL
DIVISION
RESULT
REMAINDER
(HEX)
590 / 16
36
E (14 decimal)
36 / 16
2
4 (4 decimal)
2 / 16
0
2 (2 decimal)
ANSWER
24E
11
Fortran IF Construction
The IF construction is important because it allows the program to make decisions. There are two
forms: the single IF statement, and the block-IF construction.
IF Statements.
The IF statement takes the form: if ( condition ) action. What this means is that, if the
condition in brackets is met, then the specified action is taken. The action is any single Fortran
statement. The following piece of code illustrates the most common types of condition which
may be used. A simple print statement is used as the action:
integer a,b,c,d
c
a
b
c
d
=
=
=
=
1
2
3
4
c
if
if
if
if
if
if
(a.eq.b)
(a.ne.b)
(a.gt.b)
(a.lt.b)
(a.ge.b)
(a.le.b)
print
print
print
print
print
print
*,
*,
*,
*,
*,
*,
'a
'a
'a
'a
'a
'a
= b'
not equal to b'
is greater than b'
is less than b'
is greater than OR equal to b'
is less than OR equal to b'
c
if ( (a.eq.b) .and. (c.eq.d) ) print *, 'a = b AND c = d'
if ( (a.eq.b) .or. (c.eq.d) ) print *, 'either a = b, OR c = d'
if ( .not. (a.eq.b) ) print *, 'a not equal b'
c
It should be clear what each of the above statements implies.
The Block-IF Construction
The block-IF construction is more sophisticated than the IF statement, allowing blocks of
statements to be controlled by decision making conditions. The general format is as follows:
integer a, b, c, d
c
a
b
c
d
=
=
=
=
1
2
3
4
c
if (a.eq.b) then
c
c.......Several lines of code here.
c
print *, 'a = b'
12
print *, 'hello'
c
else if (a.eq.c) then
c
c.......Several more lines of code here.c
c
print *, 'a = c'
print *, 'hello'
c
else if (a.eq.d) then
print *, 'a = d'
print *, 'hello'
else
print *, 'a not equal to b, c or d'
endif
This is a complicated example. In fact, the else statements are optional, and you may use as
many of them as you like. The important thing to note is the use of the endif statement which is
obligatory. In its simplest form, the block-IF would look something like this:
integer a, b
c
a = 1
b = 2
c
if (a.eq.b) then
c
c.......Several lines of code here.
c
print *, 'a = b'
print *, 'hello'
c
endif
It is also possible to nest block-IF's as follows:
integer a, b, c, d
c
a
b
c
d
=
=
=
=
1
2
3
4
c
if (a.eq.b) then
if (c.eq.d) then
print *, 'a = b and c = d'
print *, 'hello'
endif
endif
T
13
Modulus
The following are available from Fortran 77: ABS, AIMAG, AINT, ANINT, CMPLX, CONJG,
DBLE, DIM, DPROD, INT, MAX, MIN, MOD, NINT, REAL and SIGN.
In addition, CEILING, FLOOR and MODULO have been added to Fortran 90. Only the last one is
difficult to explain, which is most easily done with the examples from ISO (1991)
MOD
MOD
MOD
MOD
(8,5)
(-8,5)
(8,-5)
(-8,-5)
gives 3
gives -3
gives 3
gives -3
MODULO
MODULO
MODULO
MODULO
(8,5)
(-8,5)
(8,-5)
(-8,-5)
gives 3
gives 2
gives -2
gives -3
According to the book Fortran 90-95 Explained,
mod(a,p) returns a-int(a/p)*p and
modulo(a,p) returns a-floor(a/p)*p if a and p are real,
or a-floor(a//p)*p if a and p are integers,
where // is "ordinary mathematical division".
The sign of the result of mod(a,p)
follows the sign of a whereas the sign of the result of modulo(a,p)
follows the sign of p.
14