12_Y86

SP=1239H, SS=9876H, the physical address is OAAAFH O AAAFOH O 1всобн о O H66666 The instruction JMP BX is a * direct jump True O False O

Assembly language x86 using MUL

microprocessor 8086

Carider the data path below for a single cyde 32-bits MIPS processor Amume that we are ecuing the folowing instruction ADD $2, S3, Suo What is the value of the element pointed by awrow number 1 by in hexadecimal? Note that the PC and the content of registers SID and S1 are found in bottom left of the fgre below Address Content OK000016EC ON0000ABOD Data Memory OX000016FO ON0OA01245 OX000016F4 Ox00001A42 MentaFlagy Conte Ox000171C ON00OB124F Ox0001720 Ox00021345 Fead Ox0001724 OX000067AB ALU Ox0001734 OX0000AB35 meory Ox0001738 ONO000FA72 Ox000174C ON0000ABOC $s0 = OX0000AFO0 $s2 = OX00000OBA $s3 = Ox00000001 Register File and PC PC = OX000FAC04 (Before executing ADD) li li

Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01   -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]

Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01   -What is the Value of r0 and r1 after executing LDR r1, [r0, #2]   -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]

Fill the 17 dotted blocks in the 8086 memory system installed at the base address 0000OH. 1子, s) HY62256 HYE22 Ae-A4 16.-- Dy-Dy Ae-A4 Dy-D, OE WE CE DE 名 CE HYE22 HY622 Ap-A Ao-A14 Dy-D, Dy Dy OE WE CE DE WE 14 HCE 741S139 7415129 YO YO Y1 Y1 Y2 Y2 Y3 01 MAO 01 40 02A 13i-1 BHE 62A Y5 Y5 629 Y8 020 YO 17 BC %3D

Below is a list of 64-bit memory address references given as word address. Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b 0000 4 0100 8 1000 1100 1 0001 5 0101 9 1001 d 1101 2 0010 6 0110 a 1010 e 1110 3 0011 7 0111 b 1011 f 1111 Given a direct-mapped cache with 16 word blocks, what is the hit ratio? O 0.5 O 1 O 0.75 O 0.25

physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?

MOV AX, BX is an example of _____. a. Direct addressing b. Register Indirect addressing c. Register addressing d. Implied addressing

Below shows the following hexidecimal values:Address 1000: 13Address 1001: 03Address 1002: C5Address 1003: 00 Provide the format and assembly language instruction for the following hexadecimal values.

intel 8086 microprocessor

Q6: A digital computer has a memory unit with 24 bits per word. The instruction set consists of 110 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. (a) How many bits are needed for the opcode? (b) How many bits are left for the address part of the instruction? (c) What is the maximum allowable size for memory? (d) Can we extend this instruction set by adding new 10 instructions and keeping the same length of opcode part? (e) Determine the type of this instruction set: stack-based ISA or accumulator- based ISA or general purpose registers-based ISA?

8086 microprocessor

intel 8086 microprocessor

LCPU assignment Multiply the number by 1.25: A = X * 1.25X = 0x3C (direct). Then save the result to main memory using direct addressing. Example: Multiply the value by 0.75: X * 0.75 = 0.5 * X + 0.25 * X X = 100 (0x64)100 * 0.75 = 750x64 = 01100100 00110010 [Shift 1x to the right - add 0 to the left and cut off one bit from the right]00011001 [Shift 2x to the right - add 00 to the left and truncate 2 bits from the right]01001011 [Sum of previous two transactions] 01001011 = 0x4b = 75

0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h.  Add the contents of memory location 451h to AC.  Store AC to memory location 8h.  Store AC to I/O device number 8h.

0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h.  Add the contents of memory location 451h to AC.  Store AC to memory location 8h.  Store AC to I/O device number 8h.

What will be the value of ALUSrc for add instruction Inst[25–21] rs Read register 1 ALU operation Read data 1 Inst[20–16] n MemWrite Read register 2 Registers Read Write register MemtoReg Zero ALU ALU result Inst ALUSrc Read data Address data 2 Inst[15–11]| Write data Write data Data memory RegDst RegWrite 16 MemRead Sign extend Select one: а. О b. None С. 1 d. X

For the 8086 microprocessor, show the physical addresses and the contents of memory after execution of the following directives, if DS=A800H.DATA1 DW 65DATA2 DD 0A4718HDATA3 DD 6E000CHDATA4 DB 2 6 , 3 1 , 0FFh, 0 1 1 1 1B

DEBUG shows the address 807C:010F. The corresponding physical address is ______.

The first two bytes of a 4M x 16 main memory have the following hex values:   Byte 0: FFByte 1: 01   If these bytes hold a 16-bit two's complement integer, what is the actual decimal value if: a) memory is big endian:__________________ b) memory is little endian:__________________

ISA of a hypothetical CPU 1 Address Memory: Address Data - (8-bits) LOAD M 100 25 STORE M 101 90 ADD M 102 65 SUB M 103 36 MUL M 104 22 105 77 DIV M 106 89 Where: Note: all numbers are hexadecimal M-a memory address AC - accumulator Sample program: //line 1 /line 2 //line 3 //line 4 //line 5 //line 6 LOAD 100 ADD 101 STORE 106 LOAD 102 SUB 103 STORE 105 Answer the following questions based on the given information above. 1. What is the content of memory location 106 after executing line 3? 2. What is the content of memory location 105 after executing line 6? 3. Write a program segment that will multiply the content of memory location 105 with the content of AC and store the result at memory location 100. а. b. 4. For this CPU, what is the width of the program counter? (express answer in terms of bits, do not include the word "bits" in your answer)

.Memory size = 32 M address line %3D 25 bits O 35 bits O 15 bits O

8085 Assembly language.

What will be the value of ALUSrc for add instruction Read register 1 Read register 2 Registers Read Inst[25-21] | ALU operation Read data 1 | MemWrite MemtoReg Inst[20–16] t Zero ALU ALU result Inst ALUSrc Write register Read data Address data 2 Inst[15–11) Write data Data Write data memory RegDst RegWrite | MemRead 16 Sign 32 extend Select one: а. X b. 0 О с. 1 O d. None

12. What is the content value for instruction address 803? * PC 804 AC 804 АС 1CC6 PC D4CA IR 5913 IR 6910 А. OB. PC 803 PC 803 АС B980 AC D4C4 IR 6910 IR 6910 O D.

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): ADDRESS OxFFFF OXOCOE OXOCOD Ox0C0C OXOCOB OXOCOA 0x0C09 0x0000 CONTENTS 1111 1111 1111 1111 1111 1110 1101 1100 0001 1011 1100 0101 0110 0101 1000 0111 1100 0000 0100 0000 0011 0001 0101 0010 0000 1100 0000 1101 0000 0000 0000 0000 The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) What are the contents in hex of the memory location at following address in binary: 0000 1100 0000 1110? (Enter hex like the following example: Ox2A3F)

Write code on C or C++.