COM181 assignment 1: Classtest 2014
Q1
Convert 11001001 to decimal and hex
As hex we split into two 4 bit groups, 1100 and 1001, this gives C and 9 so the hex is 0xC9
As decimal = 128+64+8+1 = 20110
Q2
Convert 123 to 8 bits unsigned binary – and show all your working
Divide 123 by 2 repeatedly, keeping track of the remainder.
2 )123 = 61 R 1 Thus ‘1’ is the least significant, rightmost, bit
2) 61 = 30 R 1
2) 30 = 15 R 0
2) 15 = 7 R 1
2) 7 = 3 R 1
2) 3 = 1 R 1
2) 1 = 0 R 1
Hence the binary is 1111011. We add a leading zero to get 0111 1011
We can check our conversion by working backwards, in 0111 1011 = 64+32+16+8+2+1 =
(or alternatively, 0111 1011 is 0x7B which is 7*16 + 11 = 123)
Q3
Convert -55 to 8 bits assuming the twos complement convention is in use.
We can’t work directly with negative numbers, we need to get the binary for +55 and then take the two’s
complement – taking the two’s complement of a number, changes its sign.
5510 is 32+16+4+2+1 or
00110111
The one’s complement is
11001000 found by changing 0 -> 1 and 1 -> 0
We add one to get two’s complement
+1
So the binary for (-55) is
11001001
We can check this by adding (-128) + 64 + 8 + 1 = -55
(in two’s complement notation the sign of the leftmost multiplicative weight is negative)
Q4
Subtract 55 from 123 by adding the result of Q2 and Q3 together. Convert the binary answer back to decimal
0 1 1 1 1 0 1 1 = 123
11 11 01 01 1 01011 = -55
0 1 0 0 0 1 0 0 We can check this by noting that 123 - 55 is 68 which is 64 + 4
(The addition is done from the right, 1 and 1 is 2, which is 0 and carry 1 etc)
Given (a) is 1010 1101 0001 0000 0000 0000 0000 0010two and (b) is 1111 1111 1111 1111 1011 0011 0101 0011 two
Q5,Q6 What decimal number does (a) and (b) above represent assuming that they are two’s complement integers?
Both are negative, convert to positive, work out the decimal value and stick a ‘-‘ (minus) sign in front of them.
1010 1101 0001 0000 0000 0000 0000 0010two
0101 0010 1110 1111 1111 1111 1111 1101 + 1 = 0101 0010 1110 1111 1111 1111 1111 1110 which is 0x52EFFFFE
Which is 5*16^7 + 2*16^6 + 14*16^5 + 15*16^4 + 15*16^3 + 15*16^2 + 15*16^1 + 14*16^0
= 5*268435456 + 2*16777216 + 14*1048576 + 15*65536 + 15*4096 + 15*256 +15*16 + 14 = 1391460350 so answer
(a) is -1391460350
1111 1111 1111 1111 1011 0011 0101 0011 two
0000 0000 0000 0000 0100 1100 1010 1100 + 1 = 0000 0000 0000 0000 0100 1100 1010 1101 = 0x00004CAD
Work this out using base 16 = 4*16^3 + 12*16^2 + 10*16^1 + 13 = 4*4096 + 12*256 + 10*16 + 13 = 19629
So the answer is -19629
Given (a) is 1010 1101 0001 0000 0000 0000 0000 0010two and (b) is 1111 1111 1111 1111 1011 0011 0101 0011 two
Q7,8 What decimal number does (a) and (b) above represent assuming that they are unsigned integers?
With such big numbers it makes sense to use base 16 arithmetic (avoids dealing with numbers like 4 billion!)
1010 1101 0001 0000 0000 0000 0000 0010 is 0xAD100002
is 10*16^7 + 13*16^6 + 1*16^5 + 0*16^4 + 0*16^3 + 0*16^2 + 0*16^1 + 2*16^0
= 10*268435456 + 13*16777216 + 1048576
+ 2
= 2,903,506,946
1111 1111 1111 1111 1011 0011 0101 0011 is 0xAD100002
ie 15*16^7
+ 15*16^6
+ 15*16^5 + 15*16^4 + 11*16^3 + 3*16^2 + 5*16^1 + 3*16^0
= 4026531840 + 251658240 + 15728640 + 983040 + 45056 + 768
+ 80
+3
= 4,294,947,667
Q9,10
Give the hexadecimal representation of (a) and (b) above
0xAD100002 and 0xAD100002
Given (c) 7FFFFFF 16 and (d)
1000ten
Q11,12 Convert (c) and (d) above to 32 bits of binary
0111 1111 1111 1111 1111 1111 1111 1111 and for 1000 ten we have to convert to binary = 512+256+128+64+32+8
0000 0000 0000 0000 0000 0011 1110 1000
(0x000003E8)
Q13
Give the result in binary after performing an OR of (a) and (b) together
0111 1111 1111 1111 1111 1111 1111 1111
0000 0000 0000 0000 0000 0011 1110 1000 OR
0111 1111 1111 1111 1111 1111 1111 1111
Q14
Give the result in binary after performing an AND of (a) and (b) together
0111 1111 1111 1111 1111 1111 1111 1111
0000 0000 0000 0000 0000 0011 1110 1000 AND
0000 0000 0000 0000 0000 0011 1110 1000
Q15
Give the result in binary after performing an ADD of (a) and (b) together
0111 1111 1111 1111 1111 1111 1111 1111
0000 0000 0000 0000 0000 0011 1110 1000 +
1000 0000 0000 0000 0000 0011 1110 0111
Q16
= 0x800003E7 (confirmed by windows calculator)
Convert 1234567 to hex by dividing by 16, show your working
16)1234567
16) 77160
16) 4822
16)
301
16)
18
16)
1
= 77160 remainder 7 This is the rightmost, least significant hex digit.
= 4822 remainder 8
=
301 remainder 6
=
18 remainder 13 remember 13 is D in hex
=
1 remainder 2
=
0 remainder 1
Hence the hex is 0x12D687 (confirmed by windows calculator)
Q17
Hence give the binary for Q16 ANSWER 0x12D687 is 0001 0010 1101 0110 1000 0111
Q18
Give the binary representation for -1234567 assuming 32 bit two’s complement is used.
Need to add leading zeros to the binary answer in Q17 and then invert and add one to change its sign
0000 0000 0001 0010 1101 0110 1000 0111
1111 1111 1110 1101 0010 1001 0111 1000
+1
1111 1111 1110 1101 0010 1001 0111 1001
Assuming the IEEE 754 format for 32 bit floating point numbers is to be used, answer Q19 and 20 below. The IEEE 754
has a mantissa sign bit in b31, an 8 bit exponent in b30 to b23 biased by adding 127 to the raw value and a 24 bit
number stored in the remaining 23 bit - the number is normalised to 1.xxxx and the leading bit discarded.
Q19 Convert 1956 to 32 bits of floating point – show your working
1956 = 111 1010 0100
So as a 24 bit number normalised this is 1.11 1010 0100 000000000000 this is stored without the leading ‘1’
We need to shift 1.11 1010 0100 ten bits to get the original number 111 1010 0100 so the exponent is 10. We add
127 to 10 to get 137 and convert this to binary as 10001001 (128+8+1). Hence
Floating point is 0 10001001 11101001000000000000000
Q20
Convert 0x42280000 to binary and hence to a floating point number.
0x42280000 = 0100 0010 0010 1000 0000 0000 0000 0000
This is normalised as 1.00001000101 and needs shifted left 30 bits to get the number above. Hence the exponent is
30. Also the number is positive so the sign bit (bit 31, on the left) will be ‘0’.
The exponent has to have 127 added to it so the 30 becomes 157 as unsigned binary this is 128+16+8+4+1 =
10011101.
The floating point number is 0 10011101 00001000101000000000000 Confirmed using http://www.hschmidt.net/FloatConverter/IEEE754.html