2. Assume a computer that has 32-bit integers. Show how each of the following values would be stored sequentially in memory, starting at address 0x100, assuming each address holds one byte. Be sure to extend each value to the appropriate number of bits. You will need to add more rows (addresses) to store all five values. Byte Order a) 0xAB123456 Address Big Endian Little Endian b) 0x2BF876 0x100 c) 0x8B0A1 0x101 d) 0x1 0x102 0x103 e) 0xFEDC1234 0x104 0x105 0x106 Ox107
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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 = 0XD4Assume 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 ?Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t *q uint16_t x = (uint16_t *)p; q[0]; aa aabb bbaa aabbccdd ddccbbaa
- 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)Assume that CS=3500, DS=4500, SS=5500, SI=2200, DI=4200, BX=7300, BP=8000, AX=3420 (all values are in hex). Calculate the physical address of the memory and show the contents in each of the following: a) MOV [BP]+10,AX b) MOV [SI],AX c) MOV [BX][DI]+20,AXThe memory location at address of 0X003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001 0010), see figure below. There is a 2-byte variable M which hold binary information M (1110 0101 0000 1i11). What is the address in hexadecimal format for 2-byte memory variable M, following little Endian computer? 7 Address in Data in Hex. Format Hex. Format 0X003FBF04 1110 0101 M OX003FBF03 0000 1111 0X003FBF02 0001 0010 0X003FBF01 0010 0001 J Its address in hexadecimal is 0X003FBF02. а. Its address in hexadecimal is 0×003FBF03. O b. Its address in hexadecimal is 0X003FBF04. Its address in hexadecimal is 0×003FBF01. d.
- The following data segment starts at memory address 1000h (hexadecimal) .data printString BYTE "ASSEMBLY IS FUN",0 moreBytes BYTE 25(DUP)0 dateIssued DWORD ? dueDate DWORD ? elapsedTime Word ? What is the hexadecimal address of dueDate ? a. 1045h b. 1029h c.1010h d. 102DhThe table below shows a segment of primary memory from a Von Neumann model computer Address Data 10101000 10001000 11001000 10011001 10100000 10101010 10110100 10111011 10001100 11001100 The program counter (PC) contains a value of 11001000. Find the value (in binary) that will be placed in MAR (memory address register)? MAR (in binray) %3D Find the value (in binary) that will be placed in MBR (memory buffer register)? MBR = (binray) %3DIf we store the following array of 1-byte values at memory address 0x54, what is the memory address of the value in arr[3]? arr = [0x10, 0x42, 0x09, 0x23, 0x77, 0x92 ] A. 0x23 B. 0x54 C. 0x56 D. 0x57 O E. None of the above
- 6. Write down the three-address codes for the following code segment. S := 0, I := 0 WHILE I < 10 DO S := s + X[I] I := I + 1 ENDWHILE. Assume SP=0XE99D, R16=0XE2, R17=0x25, R01=0XFC, R15=0X1F and the following memory information. Address contents (hex) post Address contents (hex) post pre 22 pre 44 OXE996 OXE99C OXE997 46 OXE99D C5 OXE998 17 OXE99E Аб OXE999 21 OXE99F 77 ОхЕ99A F2 OXE9A0 78 OXE99B C3 OXE9A1 A5 Find the values of the registers SP, R01, R16 and R17 after the following operations. РОP R01 РО R16 РОP R17 РOP R20 PUSH R15 SP R16 R17 R01 R20 R15LCPU 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