What should I do initially to fix the damaged data problem?
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What should I do initially to fix the damaged data problem?
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- Can you check whether what i did is correct or not. Table 1 shows the content inside computer memory addresses. The code listed in Table 2 is designed to sum the numbers stored in 10 locations beginning with x3100, leaving the result in R1. The initial value of R1 is 0, R2 is x3100 and R4 is 10. Your task is to trace the content of R1, R2, and R4. Table 3 is given as a guide on how to write the contents of the registers. Draw the table in your answer booklet. Table 3 PC R1 R2 R4 x3000 0 x3100 10 x3001 0 x3100 0 x3002 0 x3100 10 x3003 0 x3100 10 x3004 0 x3100 10 x3005 0 x3101 10 x3006 x3107 x3101 10 x3007 x3107 x3101 9 x3008 x3107 x3101 9 x3009 x3107 x3101 9 [3000] 0101001001100000 - AND R1 R1 #10[3001] 0101100100100000 - AND R4 R4 #0[3002] 0001100100101010 - ADD R4 R4 #10[3003] 1110010011111100 - LEA R2 x3100[3004] 0110011010000000 - LOOP LDR R3 R2 #0[3005]…Please make a cpp code about this problem. I'll give a 5 star review after as a thanks.what's the problem ?
- Draw out how the following code is run step-by-step with freezing. What is the CPI? Add r1, r2, r3 Add r4, r5, r6 Lb r7, 0 (r1) SW 0(r4), r1 Xor r2,r7,r8 Sub r3, r2, r1Can you forsee any problems with keeping the data currentTable 1 shows the content inside computer memory addresses. The code listed in Table 2 is designed t the numbers stored in 10 locations beginning with x3100, leaving the result in R1. The initial value of R1 is 0, R2 is x3100 and R4 is 10. Your task is to trace the content of R1, R2, and R4. Table 3 is given as a guide on how to write the contents of the registers. Draw the table in your answer booklet. Table 1 Address Contents x3100 x3107 x3101 x2819 x3102 x0110 x3103 x0310 x3104 x0110 x3105 x1110 x3106 x11B1 x3107 x0019 x3108 x0007 x3109 x0004 Table 2 Address 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 x3000 0 1 0 1 0 0 1 0 0 1 1 0 0 0 0 0 x3001 0 1 0 1 1 0 0 1 0 0 1 0 0 0 0 0 x3002 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 x3003 1 1 1 0 0 1 0 0 1…
- Explain how space and time overheads arise from the use of paging, and how the Translation Lookaside Buffer (TLB) mitigates the time overheads. (DON'T COPY PASTE FROM SOMEWHERE ELSE PLEASE).One reason the assembler's object files are not executable is that each object file is created independently of the others in the programme. As a result, they are unaware of the existence of other object files in the programme; linking object files is required to produce an executable file. Can an object file be run if it contains only one source file and does not include any library files? What are your reasons for or against?Have you considered the possibility of an interruption? Where is the relevance to context flipping in this?