QI) I BX=1000, DS-0200, SS=0100, CS-0300 and AL=EDH, for the following instruction: MOV [BX] + 1234H, AI. Find the physical address in the memory. Q2) For the above question, if BP is used instead of BX, what is the physical address in the memory?
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A: Solution:-
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Q: Q:find the actual address for the following instruction assume X=A6 and ?=PC=8B79, LOAD X(PC), D…
A: The given data is. X = A6 PC = 8B79 The given instruction is: LOAD X(PC), D
Q: Q:find the actual address for the following instruction assume X=A6
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Q: Q:find the actual address for the ..il following instruction assume X=38 and R index=DDCE8 hex LOAD…
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A: The MIIPS program is given below:-
Q: Q:find the actual address for the following instruction assume X=38 and R index=DCE8 hex LOAD X(Ri),…
A: Solution:-
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Q: Q:find the actual address for the following instruction assume X=A6 and PC=8B79, LOAD X(PC), D…
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Q: Q:find the actual address for the following instruction assume X=38 and ?=R index=DCE8 LOAD X(Ri), A…
A: Given data: R index = DCE8 X = 38 Now find the actual address. The instruction is: LOAD X(Ri), A
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- Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4Design a memory map to work with 8085 Microprocessor to have 8K byte ROM and 2K byte RAM. ROM should start from memory location 0000H and RAM immediately follows it. Use exhaustive decoding scheme?Select the WRONG statement relating to the properties of DRAM memories. (Select the WRONG statement; the statements not selected should be true) The bit line is kept at a voltage between high voltage and ground, during a WRITE operation. A DRAM row-read takes a long time, but once a row is read into a buffer, multiple back-to- back reads to the same row are much faster. Unlike SRAM reads, DRAM reads are "destructive". Memory locations not accessed for some time need to be refreshed periodically.
- 3. Calculate the physical memory location for each of the following cases? a- The logical address D470H in the extra segment. b- The logical address 2D90H in the stack segment. C- MOV [BP],AL if BP=2C30H. Assume ES=52B9, SS=5D27, DS=E000, and CS=B3FF.Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Page Table Physical Memory Physical Address (starting) Oxppppdddd Page Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 PPpp: page number dddd: page offset 1 1 Оxd000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x00011119 What is the physical address for 0x00000001 What is the logical address for Oxd0000001 ? What is the logical address for Oxc0000002 ?Given the image below, what is the Linear Address of the memory location currently being pointed to in the Data Segment . Make all hexadecimal letters lower case ● Use no spaces in your answer You may proceed the answer by Ox if you like (lower case x) Do not include leading zeros (in the most significant nibbles) in your answer Segment Descriptor Table ??? 19 0x1A0020 0x0224 20 0x2B0020 ??? 21 0x000F5B OXOA 22 0x09223A 23 0x09223A CS 0022 DS 0019
- funCount: MOVLW .15 MOVWF Ox82 count: DECF Ox82 BNZ count RETURN Assuming Oscillator frequency is set to 4 Mhz, fill in the following blanks based on the execution of the above code segment: Instruction cycle period, Tinst = uSec Number of instruction cycles required to execute the code Total time to execute the code uSecSelect all true statements. Segmentation always uses 32-bit logical addresses. The memory management unit utilizes the segment component of the logical address to get the segment table start address and adds the offset to obtain the physical address. Segmentation uses segment and offset logical addresses. Maximum segment number limits segment length. Segmentation restricts process memory access to respective segments. Segments may be granted privileges.Find the PA of the memory location and its contents after the execution of the following assuming that DS=1512h. MOVAL, 99H MOV [3518], AL
- Part 1: Preliminary Summarize the difference between binary machine language, assembly language, and high-level source files like C. Contrast the difference between a compiler, assembler, and linker What is the difference between the data segment and the text segment of in an assembly language program? What register is used as the stack pointer? What is a ebreak instruction in the sample program? How would you define a constant named 'BLUE' and assign it to the value OX00F in assembly language? Describe how to implement the psuedo-instruction li t1, e using a native instruction.Suppose a computer using direct mapped cache has 224 bytes of byte- addressable main memory and a cache size of 64K bytes, and each cache block contains 32 bytes. (Note: 64K = 26 * 210) a) How many blocks of main memory are there? b) What is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields?Below is an explanation of the differences between using direct and indirect addressing. Give an instance.