Three Instructions and One ValueHere are three segments of a program.After the LC-3 executes the three instructions starting at address x4810 a value is put into register R6.x4010 1110110000101111x4011 011011e110111111x4012 011011e110111111X403F 01100ee010110011x4040 01100ee010110011x4041 01100e001011e010X60B2 11001e1011111110X60B3 1011111010101011X60B4 1011111010101101The hex (with an x) value in register R6:

Three Instructions and One Value Here are three segments of a program. After the LC-3 executes the three instructions starting at address x401e a value is put into register R6. x4010 11101100e0101111 x4011 0110110110111111 x4012 011011e110111111 X403F 01100ee010110011 x4040 0110eee01011ee11 x4041 011000001011e010 X60B2 11001e1011111110 X60B3 1011111010101e11 X60B4 1011111010101101 The hex (with an x) value in register R6: ..

Three Instructions and One Value Here are three segments of a program. After the LC-3 executes the three instructions starting at address x401e a value is put into register R6. x4010 11101100e0101111 x4011 0110110110111111 x4012 011011e110111111 X403F 01100ee010110011 x4040 0110eee01011ee11 x4041 011000001011e010 X60B2 11001e1011111110 X60B3 1011111010101e11 X60B4 1011111010101101 The hex (with an x) value in register R6: ..

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

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?
microprocessor 8086
If R0 = 0x20008000, after STMDA r0!, {r3, r9, r7, r1, r2} instruction is executed, register r7 will be stored in memory starting from which memory base address. A. R0 = 0x20007ff0 B. R0=0x20007fec C. R0 = 0x20007fff D. R0= 0x20007ff4 E. R0 = 0x20007ffef
Two word-wide unsigned integers are stored at the memory addresses 0A00 H and 0A02 H respectively, Write an instruction sequence that computes and stores their sum, difference, product, and quotient then Store these results at consecutive memory locations starting at address 0A10 H in memory. To obtain the difference, subtract the integer at 0A02 H from the integer at 0A00 H. For the division, divide the integer at 0A00H by the integer at 0A02 H. Use register indirect relative addressing mode to store the various result.
For the instructions below, assume the initial conditions shown for r3-r5 and the flags. Show the value of any registers that are changed. If the instruction affects flags, show the value for all affected flags and explain the reason. Initial conditions: r3 =0xF0E1_D2B3 r4 =0x8406_3F40 r5 = 0xA7F0_8008 NZCV = 1010 Instruction: ANDS r3,r4,r5 Edit View Insert Format Tools Table 12pt ✓ Paragraph ⠀ Decimal 0 1 2 3456789 10 11 45 12 с 13 D E F 14 Hex 0 1 2 3 4 5 6 15 7 8 9 A Binary 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
The addition or subtraction must be performed in HEX or BIN and can NOT be performed in DEC since those numbers represent signed OR unsigned values (These instructions are NOT executed one after the other one; instead, each instruction starts with the initial conditions given in the statement.) (a) 0x92394103 + 0xA04230C8 (b) 0x12345678 – 0xEDCBA988
Fill in the following table showing the results of the indicated instructions, using the register values shown. Also, show the OF and CF values AFTER the instruction executes. Show all results in hexadecimal. Show OF and CF as 0 or 1. Instruction %cl %al Result (in hex) OF CF 1. addb %al, %cl Ox52 Ox4D 2. addb %al, %cl Ох54 Ох67 3. addb %al, %cl OX9A OXD5 4. addb %al, %cl Ох95 OXBE 5. addb %al, %cl OXC4 Оx47 6. addb %al, %cl Ox9F OX7E 7. addb %al, %cl Ox24 Ox5B 8. addb %al, %cl Ox45 Ox3
Use the Fetch Decode Execution steps to execute the assembly language into machine instructionsand 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 Mnemonic table Op-code Mnemonic 001 LOAD 010 STORE 011 ADD 100 SUB 101 EQUAL 110 JUMP 111 HALT
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 00000002 00000007 0000000A 0000000C
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
Please help me with this computer architecture homework
Tompute the physical address for the specified operand in each of the following instructions. egister contents and variable are as follows: (CS)=0A00H, (DS)=OB0OH, (SS)=OD00H, S)=OFFOH, (DI)=00BOH, (BP)=00EAH and (UP)=0000H, LIST=00FOH, AX=4020H, BX=2500H. 1) Destination operand of the instruction MOV LIST (BP+DI], AX 2) Source operand of the instruction MOV CL, [BX+200H] 3) Destination operand of the instruction MOV [DI+6400H], DX 4) Source operand of the instruction MOV AL, [BP+SI-400H] 5) Destination operand of the instruction MOV [DI+SP], AX 6) Source operand of the instruction MOV CL, [SP+200H] 7) Destination operand of the instruction MOV [BX+DI+6400H], CX 8) Source operand of the instruction MOV AL , [BP-0200H] 9) Destination operand of the instruction MOV [SI], AX 10) Destination operand of the instruction MOV [BX][DI]+0400H,AL 11) Source operand of the instruction MOV AX, [BP+200H] 12) Source operand of the instruction MOV AL, [SI-0100H] 13) Destination operand of the…