Question

ARM assembly language

Write the final updated values for each memory and register after the given instruction executes in the space provided.There are multiple parts of this problem. Note Only write the change values if the values didn't change. Leave the updated cell blank.

please explain (as simple as possible ) whats going on after the given command is executed .

Q.5.4 Bring to class LSL R6, R6, #3 Updated Data Updated Data Memory Address Ox84F0 Ox 841 Ox84F2 Ox843 Ox8454 Ox84F5 Ox84F6 Ox84F7 Ox8458 0x8459 Ox84FA Ox84FB Ox84FC OxB4FD Ox84FE Ox84FF Ox8500 Ox8501 Ox8502 0x8503 Ox8504 Ox8505 0x8506 Ox8507 Ox8508 Ox8509 Ox850A Ox850B Ox850C 0x850D Ox850E Ox850F Orig. Data Ox82 Ox32 OxBC OxA4 OxF4 Ox6A Ox9D Oxc2 Ox42 Ox67 Ox8D OxA1 OxF2 OxB8 Ox5E OXEE OxFA OxF3 Ox54 OxBA 0x6D Ox9C Ox78 OxBB OxC3 OxB4 OxAA OxD9 OxDA OXFF OXES Ox33 Register RO R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 R12 R13 R14 R15 Orig. Data OxDC002B45 Ox00008500 OxAB038730 0x2B78DFEO Ox000084F0 0x00100000 Ox041 BOCOD Ox00100000 0x00000000 Ox00842ABO OxOABD9830 Ox0000000A Ox00000002 OX00008500 Ox003762 DD Ox00100000 Flags N Z с V Q 0 1 1 0 0

Answer 1

LSL - Logical Shift Left

R6 = 0x041B0C0D = 0000_0100_0001_1011_0000_1100_0000_1101

After execution of LSL R6, R6, #3

R6 = 0010_0000_1101_1000_0110_0000_0110_1000 is shifted left by 3 bits position and 0's enters from left.

After execution of above instruction, only R6 contents will change.

Therefore R6 = 0x20D86068