Given the following memory addresses and opcodes to be executed, if the current value of the Instruction Register (IR) is EBOA, what is the value of the Instruction Pointer Register (RIP) after the instruction in IR is executed? 00000000 EBOA 00000002 B805000000 00000007 BF01000000 0000000C BAODO00000
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- Given the following branch instruction and location, answer the following questions about it. Address Instruction Ox10000 Oxb5000184 What type of Branch instruction is this (CBZ, CBNZ, B, etc.): What is the offset in instructions of the branch target (write as a hex number with "-" if negative): What is the offset in bytes of the branch target (write as a hex number with "-" if negative): What is the address of the branch target:Given following code: 00 9000 01 1150 02 1251 03 5300 04 6420 05 7541 06 8414 07 D20A 08 2523 09 0000 0A 3331 0B B200 0C D20A 0D 2523 0E 0000 50 000A 51 0005 52 0000 Explain what each instruction does in the program, stored in the memory locatons 00 through 0E. When the program stops, what are values in registers R0, R1, R2, R3, R4, R5, and in memory location 52? What does the program do?if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical address is O 6243H O 4234H O 6234H O 6324H O 4244H
- Convert the given hex value to its R-format MIPS instruction by completing the following table. The instruction and letters for registers must be in lower case. DO NOT TYPE DOLLAR SIGNS WITH REGISTERS. Hex Opcode (decimal) rs (decimal) rt (decimal) rd (decimal) funct (decimal) Instruction 02488822 O 18 16 17 34 sub s4 $$1 " $ to " LAA digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 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 ofmemory.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. What is the largest unsigned binary number that can be accommodated in one word of memory?Suppose that you have a computer with a memory unit of 24 bits per word. In this computer, the assembly program's instruction set consists of 198 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. How many additional instructions could be added to this instruction set without exceeding the assigned number of bits? Discuss and show your calculations.
- SP=1239H, SS=9876H, the physical address is AAAFOH Non of them 1BC06H 0AAAFH 99999H if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical address is 6324H O 4244H 4234H 6234H 6243H OUse the Fetch Decode Execution steps to execute the assembly language into machine instructions and execute the instruction as the computer would. Explain into detail what happens during the execution of each statement. # Machine Code 0 001 1 111100 1 010 0 110000 2 001 1 101000 3 010 0 110111 4 001 0 110000 5 011 0 110111 6 010 0 111000 7 111 0 000000 Mnemonics Table Op-Code Mnemonics 001 LOAD 010 STORE 011 ADD 100 SUB 101 EQUAL 110 JAMP 111 HALTIf DS=7F0OH, BX=37EH and AX=1F05H, find the value of (AL) after executing (XLAT) instruction, if you have the following memory contents:* 7AH 7F386H 66H 7F38SH BOH 7F384H 79H 7F383H F4H 7F382H 8CH 7F381H F4H 7AH 79H 66H BOH 8CH What is the missing instruction in the following code if you are required to calculate (A7H/15H) ? - MOV Bх, 15H Mον AX , A7Η Div BX CWD MOV AH, O MOV BX, AX MOV DX, AX MOV DX,0 CBW XLAT O O O O O
- What is the effective address that is targeted by the store instruction whose code word in binary is:101011 01000 10001 1111 1111 1111 1000Assume [$t0]=0x400CYou are given a before condition of the registers eax, ebx, edx and an instruction. Choose the correct after condition of eax, edx Before condition: EAX: 00 00 00 11 EBX: 00 00 00 OC EDX FF FF FF 03 Instruction: CWD idiv bx After condition: EAX: 00 00 00 51 EDX: FF FF FF 01 O EAX: 00 00 05 01 EDX: FF FF FF 03 EAX: 00 00 00 01 EDX: 00 00 00 05 O EAX: 00 00 00 01 EDX: FF FF 00 05Bus d'adresse Ox0 A7 0 A6 0 AS Oxo 0 A4 0 A3 0 A2 0 A1 Lecture Bus de contrôle Écriture AO Bus de données Mémoire d'Instructions Adr Adr Ctr Ctr Ctr Ox Mémoire de données Clavier Écran CD 0 EN Data Data EN Data EN Data EN Write the instruction STR R2, [R3] in hexadecimal (using 2 bytes). АО A1 DO D1 D d u r D3 D2 09