Question
Need help with this Assume that memory contexts/data are constantly RECYCLED every 8 memory single words, hence Mem[32] = Mem[0], Mem[36] = Mem[4], Mem[40] =
Need help with this
Assume that memory contexts/data are constantly RECYCLED every 8 memory single words, hence Mem[32] = Mem[0], Mem[36] = Mem[4], Mem[40] = Mem[8], ... Memory addresses start strictly from 0 and grow ONLY POSITIVE. If a negative address is calculated then the Operating System throws an exception and the program is TERMINATED. Also, the low address of memory lanes start from LEFT to RIGHT side (check below).
General question for guidance: Show the contents of memory entries that have changed as well as the value stored in x1, x2, x5, x6 after running this piece of code below. Show the values in HEX.
Task 2 : x20 contains number (in hex): 09A008C09DD9E8084. Memory illustration below does not contain this number. You must place it there at the start of your handling the code. Consider the Type of Endianness.
Task 3: At the line of code indicated, you must add number 0x5F 34 6E B9 89 22 29 A1 to register x3, using only real RISC-V instructions (no pseudo instructions allowed).
Task 4: Start by assuming the machine is: 1) Little Endian, 2) Big Endian. If you find that one type of Endianness or both lead at any point to negative memory access, document where this happens. Then, switch to the other Endianness type and recalculate those instructions and tasks that are affected. One of the two types of Endianness will lead to the programs completion. .
sd x20, 7(x4) /write this to memory --- it is not placed in the scheme below... // TASK 2: change the memory content by executing the instruction above - write to memory lh x11, 24(89) // offset is in decimal number, careful: check how lh works // must distinguish between LE and BE addi X2, x11,6 // Line A // TASK 3: Write your code below, using the least costly REAL instructions: // Line B *** add number: 0x5F 34 6E B9 89 22 29 A1 to register x3 - same for LE-BE srai x1, x3,8 ld x5, 6(x2) // Line C -- must distinguish between LE and BE xor x6, x5, x1 sh x6,-3(x2) // Line D -- must distinguish between LE and BE Mem Low Order Content Address (decimal) 28 12FF0120 24 21004204 20 18926163 16 7E1565A9 83E5CFC6 8 D4CB3F59 4 7C32018E 0 F3B946EC 12
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