ithout using LXI command how you will transfer one byte from memory location whose address is 11H to another memory location whose address is 13H( 8085 SIMULATOR
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without using LXI command how you will transfer one byte from memory location whose address is 11H to another memory location whose address is 13H( 8085 SIMULATOR ).
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- write the code in program in 8085 SIMULATOR : with using LXI command how you will transfer one byte from memory location whose address is 11H to another memory location whose address is 13H.write in program in 8085 SIMULATOR : with using LXI command how you will transfer one byte from memory location whose address is 11H to another memory location whose address is 13H.Physical address is formed from segment address & offset address. Select one: O True O False
- 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 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]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]physcal 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?
- . 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 = 75Difference between logical and physical addresses is in:
- According to the memory view given below, if RO = 0x20008000, then LDRSB r1, r1 = ?(data overlay big endian)? Memory address Data Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 ØxC3 ØX20007FFE ØXD4 ØX20007FFE OXE5 (Ctrl) A-R1 = 0XC3 B-R1 = 0x000000C3 C-R1 = OXC3000000 D-R1 = 0xffffffC3 E-R1 = OxC3ffffffyou have a memory of 128 gbs, which is divided into 16 blocks. Give the complete addresses of the following bytes 8th byte in 15th block5th byte in 11th block7th byte in 9th blockadres 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