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- On a typical microprocessor, a distinct I/O address is used to refer to the I/O data reg- isters and a distinct address for the control and status registers in an I/O controller for a given device. Such registers are referred to as ports. In the Intel 8088, two I/O in- struction formats are used. In one format, the 8-bit opcode specifies an I/O operation; this is followed by an 8-bit port address. Other I/O opcodes imply that the port ad- dress is in the 16-bit DX register. How many ports can the 8088 address in each I/O addressing mode? .Let, in 8086 based system current SP value is FFFAh and the CS, IP and FL values are 7000h, 1000h and 1001h, respectively. Now, if at that instance an interrupt occurs, then show the stack increase and decrease scenario along with changes in SP values and stack contents.a. Find the address accessed by each of the following instructions. If DS = 0100H, BX= 0120H, DATA = 0140H, and SI = 0050H and real mode operation:1. MOV DATA[SI], ECX2. MOV BL, [ BX+SI]b. Descriptor contains a base address of 00260000H, a limit of 00110H, and G = 1,determine starting and ending locations are addressed by the descriptor for aCore2.
- Question 4: There is an application that requires the hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables, TWO Data RAMs of 8Kx8. The memory map of the design should be: Program ROM should start at address 0000μ. Then, the Data ROM should come above the Program ROM. Finally the Data RAMs must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs or RAMS. A. Using logic gates, draw the pin connections of the design. Label your diagram fully. B. Calculate the address space of the ROMs and RAMs of your design. C. Show the design's address space on a memory map, starting with 0000μ at the bottom and FFFFH at the top.In a system the contents of PC, Base register, Register R1, and Register R2 has contents 80, 140, 20, and 40, respectively. Content of R2 is used as the displacement value. What is the effective address computed using the Base addressing and the Relative - Base addressing modes, respectively?LDR uses an address called a "Register Offset" address. Why? Group of answer choices The data value loaded is at an address in the memory. The data value loaded is the results of adding a register and immediate to obtain the reference for the data value Memory is accessed at a location that is a numeric value, starting at (offset from) 0 The ALU can only use register (not memory) values All data access is relative to a memory address or 0. The LDR command uses what units in the CPU? Select all that apply. Group of answer choices Data Memory Multiplication Unit Barrel shifter ALU Addition is done in what CPU unit? Group of answer choices The multiplication unit The registers The ALU The barrel shifter
- perform early and late binding and explain what happens when we perform early and late binding in context of address.Write a service routine which resets all elements of an array that resides in memory location from A000 H to A0FF H with DS equal to 0000 H. The service routine address is CS:IP where CS is 2000 H and IP is 0100H. Assume the interrupt type that is called is 50 (x8086- nano)Problem 1. This problem is about operand modes, in particular about memory addressing using the operand modes described in lecture and the textbook. The following shows the contents of a portion of the program memory (locations Ox10000 through 0x10040), and certain registers. All contents are 64-bit quantities. Address Memory Contents 0x10040 0x10038 0x10030 0x10028 0x10020 0x10018 0x10010 0x10008 0x10000 a. movq (%rdi), %rax b. movq (%rdi,%rsi), %rax 100 0x10040 200 25 0x10028 0x10020 c. movq 8(%rdi, %rcx, 4) %r8 d. movq -8(%rdi,%rsi), rbx 500 0x10018 2 Register Contents %rdi 0x10008 %rsi %rdx %rcx For each of the following instructions, say what value ends up in the destination register. Each instruction starts with the state shown above, i.e., the effects do not accumulate. 0x20 8 4
- Evaluate the expression: F = (c -a)*d +e-b focusing 1- address through 3-address format. Also explain which addressing mechanisms bears maximum number of instructions and why?In [COOK82], the author proposes that the PC-relative addressing modes be elimi- nated in favor of other modes, such as the use of a stack. What is the disadvantage of this proposal?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…