the hexadecimal address of dueDate ?
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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. 102Dh
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- . 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 R15Memory 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 = 0XD4A CPU that supports little endian format reads two integer (4-byte) values from address 0x1000 and 0x2000. The values read are 55 and 6850 respectively. Please show the memory contents (byte-wise) at address 0x1000 and 0x2000?
- 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 ?The following data segment starts at memory address 0x4100 (hexadecimal).data printString BYTE "Do not add decimal to hex",0 someBytes WORD 36 DUP(0) moreBytes BYTE 10, 20, 30, 40, 50, 60, 70, 80, 90 questionAddr DWORD ? ignoreMe WORD? Ans:-adres veri 01h 5x9 02h 5x8 8715683 b190100564 7156 08h 6715988 5x2 bis010054 71 09h 5x1 Write the asm code that will create the address and contents given in the table. (It is mandatory to use loop and indirect addressing.) b10100564 159 190100564-7I5
- Fill all the information to reflect the addresses of the program memory Adress code, .org 0x540 LDS R10, 0x700 LDI R20, 20 LDI R16, 10 ADD R16, R20 STS 0x700, R16 ADD R16, R10 STS 0x701, R16 Ox540 Ox541 Ox542 Ox543 Ox544| Ox545 0x546|| Ox547 Ox548 0x549 0x54ASuppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What is the Value of r0 and r1 after executing LDR r1, [r0, #2] -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]The memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.
- Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]Given rax = 0x0000000200000100, rbx = 0x0000000000000100, and rcx = 0x0000000000000001,and the following values in memoryaddress -> byte at that address0x0000000000000100 -> 0x010x0000000000000101 -> 0x000x0000000000000102 -> 0x000x0000000000000103 -> 0x000x0000000000000104 -> 0x020x0000000000000105 -> 0x000x0000000000000106 -> 0x000x0000000000000107 -> 0x00what is the new value in %rax after the following operation?subq -0x04(%rbx, %rcx, 4), %raxphyscal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?