For the code sequence below, state whether it must stall, can avoid stalls using only forwarding, or can execute without stalling or forwarding. i1: addi $t1, $to, 1 12: addi $t2, $t1, 2 13: addi $t3, $t0, 2 14: addi $t4, $to, 4 i5: addi $t4, $t3, -1 16: beq $t0, $0, end end: Which lines in the program would require stalls based on data dependencies

Dear Wrier it is my solution could oyu please double check this solution and confirm that your solution is correct one? The SLIP frame begins with the character 0xC0, known as the SLIP END character. Bytes in the frame are treated as either control characters or non-control characters. Non-control characters are sent as is, while control characters are replaced by a two-byte sequence. Specifically, 0x7E and 0xDD are non-control characters and are sent as is. 0xDB, 0xC0, and 0x7D are control characters and are replaced by the two-byte sequences 0xDB 0xDC, 0xDB 0xDC, and 0xDB 0xDD, respectively. The byte 0xDC is also a control character and is replaced by the two-byte sequence 0xDB 0xDD. Finally, the frame ends with the SLIP END character (0xC0). The bytes sent on the wire (after framing) are 0xC0 7E DB DD 7E C0 7D DB DC FF DB DD 0xC0. The frame begins with a flag sequence represented by the bit sequence 01111110 and denoted by the value 0x7E. Each byte in the frame undergoes encoding…

Problem 4: Write and run a program to find the values of each destination operand: .code mov al,01101111b and al,00101101b mov al, 6Dh and al, 4Ah mov al,00001111b or al, 61h mov al,94h xor al,37h ; a. Al= ? ; b. Al= ? ; c. Al= ? ; d. Al= ?

Choosen answer is wrong and there is no other part to the question; it is the only one.

These are using Bitwise operation in C. Can you help?