Question | Answer |
---|---|
6 | at the address 0x10010000 is 0x11 (decimal 17) / hexadecimal base / 0x00000017 is at 0x10010004 / at the address 0x10010008 is 0x0 |
7 | R3 = 0 and R4 = 0 |
9 | R3 = 11 and R4 = 28 |
10 | 0x10010008 changed to 0x28 |
12 | at the address is the liu $1, 4097 |
13 | add $4, $4, $3 is at 0x00400030 / brakepoint is at 0x00400034 |
14 | R3 = 0, R4 = 1, PC = 00400020 |
15 | PC = 00400030, R3 = 11, R4 = 17 |
16 | PC = 00400034, R3 = 11, R4 = 28 |
19 | at the memory addresses 0x10010000 - 0x10010010 are stored 0x11, 0x17, 0x28, 0x0, 0x0 |