Representation of numbers and characters
- 1. A solution manual to Assembly language Programming and the organization of the IBM PC Chapter 2 Representation of numbers and characters BY:- Ytha Yu & Charles Marut prepared by : Warda Aziz Wardaaziz555@gmail.com
- 2. Question 1: In many applications , it saves time to memorize the conversions among small binary, decimal, and hex numbers. Without referring to the table , fill in the blanks in the following table binary decimal hex 1001 9 9 1010 10 A 1101 13 D 1100 12 C 1110 14 E 1011 11 B Question 2: Convert the following numbers into decimal: A) 1110 B) 100101011101 C) 46Ah D) FAE2Ch Solution: 1110: =1*23 + 1*22 + 1*21 + 0*20 =8 + 4 + 2 + 0 =14 100101011101: =(1*211 )+(0*210 )+(0*29 )+(1*28 )+(0*27 )+(1*26 )+(0*25 )+(1*24 )+(1*23 )+(1*22 )+(0 *21 ) +(1*2 0 ) =2048 + 0 + 0 + 256 + 0 + 64 + 0 + 16 + 8 + 4 + 0 +1 2397 46Ah: =4*162 + 6*161 + A*160 =4*256+ 6*16 +10*1 =1024 + 96 +10 =1130 FAE2Ch: =F*164 + A* 163 + E*162 + 2*161 +C*160 =15*65536 + 10* 4096 + 14*256 +2*16 +12*1 =983040 + 40960 +3584 +32+12 =1027628 Question 3: Convert the following decimal numbers: a) 97 to binary b) 627 to binary
- 3. c) 921 to hex d) 6120 to hex Solution: 97 to binary ;LCM with 2, remaining bits are in binary =1110 0001 627 to binary =0010 0111 0011 921 to hex LCM with 16, remaining digits are in hex form =399 6120 to hex =17E8 Question 4: Convert the following numbers: a) 1001011 to hex b) 1001010110101110 to hex c) A2Ch to binary d) B34Dh to binary Solution: a) 1001011 to hex 1001011=0100 1011(byte complete) As we know 0100=4 and 1011=B =4B b) 1001010110101110 to hex =95AE c) A2Ch to binary A=10d=1010(binary) 2=0010(binary) C=12d=1100(binary) =1010 0010 1100b d) B34Dh to binary B=11(decimal)=1011(binary), 3=0011(binary) 4=0100(binary) D=1101(binary) =1011 0011 0100 1101b Question 5: Perform the following addition: a. 100101b + 10111b b. 100111101b + 10001111001b c. B23CDh + 17912h d. FEFFEh + FBCADh Solution:
- 4. 100101b + 10111b 100101 +10111 111100 100111101b + 10001111001b 100111101 + 10001111001 10110110110 B23CDh + 17912h B23CD +17912 C9CDF FEFFEh + FBCADh Solving from the right most values i.e., Eh+Dh=14d+13d=27d 27d=1Bh(solving through LCM method) B is written below and 1 is given carry to the left bit. Process goes so on and we get; FEFFE +FBCAD 1FACAB Question 6: Perform the following subtraction: a) 11011 - 10110 b) 10000101 - 111011 c) 5FC12h - 3ABD1h d) F001Eh - 1FF3Fh Solution: 11011 - 10110 11011 - 10110 00101 10000101 - 111011 10000101 - 111011 1001010 5FC12h - 3ABD1h ;borrow carries 0101 1111 1100 0001 0010 (5FC12h) - 0011 1010 1011 1101 0001 (3ABD1) 0010 0101 0000 0100 0001 (binary) =25041(hex) F001Eh - 1FF3Fh 1111 0000 0000 0001 1110 - 0001 1111 1111 0011 1111 1101 0000 0000 1101 1111
- 5. =D00DF Question 7: Give the 16 bit representation of each of the following decimal integers . write answers in hex. a. 234d b. -16d c. 31634d d. -32216d Solution: 234 Bin= 0000 0000 1110 1010 Hex=00EAh -16 Bin=0000 0000 0001 0000 2’s compliment: 1111 1111 1110 1111 + 1 1111 1111 1111 0000 =FFF0h Hex=FFF0h 31634 bin= 0111 1011 1001 0010b Hex=7B92h -32216 Same as Q7(part e) Bin=1000 0010 0010 1000b Hex=8228h Question 8: Do the following binary and hex subtractions by two’s compliment addition. a) 10110100b - 10010111b b) 10001011b - 11110111b c) FE0Fh - 12ABh d) 1ABCh - B3EAh Solution: Note: if the resulting bit has a carry 1 , neglect the carry if the result has no carry take 2’s compliment again with a negative sign. 10110100 - 10010111 Two’s compliment= 01101000+1=01101001 10110100 + 01101001
- 6. 1 00011101 Dropping the resulting carry bit we get the answer 0001 1101b or 11101b 10001011 - 11110111 2’s compliment 11110111=00001000+1=00001001 1000 1011 +0000 1001 1001 0100 Taking 2’s compliment again and put a negative sign 01101011 +1 01101100 Ans -1101100 FE0Fh - 12ABh FE0Fh - 12ABh=1111111000001111b-1001010101011b Take 2’s compliment of 0001 0010 1010 1011b; 1110 1101 0101 0100b+1b=1110 1101 0101 0101b 1111 1110 0000 1111 +1110 1101 0101 0101 1 1110 1011 0110 0100 Ans: =1110 1011 0110 0100b =EB64h 1ABCh - B3EAh Same as Q8(3,2) Ans: -992Eh Question 9: Give the signed and unsigned interpretations of each of the following 16 bit or 8 bit numbers. a. 7FFEh b. 8543h c. FEh d. 7Fh Solution: 7FFEh Signed: 32766 ; as the MSB is 0(in binary) Unsigned: 32766 8543h Signed: -31421 ; as the MSB is 1 Unsigned: 34115 FEh Signed: -2 ; as the MSB is 1 (see table 2.4B) Unsigned: 254 7Fh Signed: 127 ; as the MSB is 0
- 7. Unsigned: 127 Question 10: Show how the decimal integer -120 would be represented a) In 16 bits b) In 8 bits As we know 120d=0111 1000(binary) -120=2’s compliment ANS: 16 bit: (0000 0000 1000 0111)16 + 1= (1111 1111 1000 1000)16 8 bit: (1000 0111)8 + 1= (1000 1000)8 ;2’s compliment because we are solving for negative value i.e., -120 Question 11: For each of the following numbers ,tell whether it could be stored (a)as a 16 bit number or (b) as an 8 bit number? a. 32767 b. -40000 c. 65536 d. 257 e. -128 ANS: Number Bit storage 32767 16 bit -40000 Bigger than 16 bit 65536 Bigger than 16 bit 257 16 bit -128 both Question 12: For each of the following 16 bit numbers, tell whether it is positive or negative? ANS: Signed number polarity 1010010010010100b negative 783E3h positive CB33h negative 807Fh negative 9AC4h negative Question 13:
- 8. If the character string “$12.75” is being stored in memory starting at address 0, give the hex contents of byte 0-5. Data string Data $ $ 24 1 1 31 2 2 32 . . 2E 7 7 37 5 5 35 ANS: The content at address 5 is 35. Question 14: Translate the message of ASCII 41 74 74 61 63 6B 20 61 74 20 44 61 77 6E ANS: Attack at Dawn Question 15: Suppose that a byte contains the ASCII code of an upper case letter. What hex number should be added to it to convert it to lower case? ANS: 20h Question 16: Suppose a byte contains a decimal digit that is “0”….“9” . what hex number should be subtracted from the byte to convert it to the numerical form of chracters? Ans:30h