Guilty Crown – 10

Sorry, finals happened. One per day, four days in a row, somewhat.

This is the question which made me go crazy. Do you think you’re man enough to solve it? [you need to know MIPS & C]. Also, assume mystery resides in 0xF0000000 if that even matters. And ignore the unsigned declaration of mystery. Mystery call in main() is calling that MIPS function. Tell me if you get the answer!

Back to GC, I feel like it’s a bit too rushed… Everything spilling over in one episode? What?

Download : [ Torrent | ASS ]

14 Responses to “Guilty Crown – 10”

  1. 1 ABC

    so, since you decided to come back to do one series this season, how about next season?

  2. 2 R

    For the life of me, I cannot imagine why anyone would want to call assembly from C.

  3. 3 omebbqwtfzombielol

    thank you and yes winter break is nice

  4. 4 huh?

    wat is dis. do i care? just gimme .ass

    Fail trolling aside, thanks for the sub. And finals will always be a bother since I just recently finished mine as well, haha

  5. 5 VanDay

    Thanx !

  6. 6 Justinnnnnn

    First time posting here…Just wanted to say a big THANK YOU for such quality releases and so fast too.

    Have a nice winter break!

  7. 7 Pie

    The answer is pie. It will always and forever be pie.

    And thanks for the release!

  8. 8 Jay

    Yeah MIPS and comp arch were annoying, but my CA professor graded easy. For the record your question crashed my mobile browser.

  9. 9 alex

    I can tell you what’s going on in the mips code but I don’t understand what the question even is so I’m afraid I can’t tell you about that. You probably already know this stuff but meh

    First it copies @ra to $t0. The jal instruction is storing the return address in $ra, overwriting the old value of $ra. But the jump return command is jumping back to $to, which is the OLD $ra. I don’t know how to call mips from C, so I don’t know what that is initially.

    At any rate, that’s the initial call. I don’t know how to parse the rest because again, I don’t know how C interacts with mips. But it shouldn’t be too hard to trace through. Just obnoxious as hell.

    I remember learning mips last semester. I’m kind of surprised to learn other people have to learn it as well – we all thought it was a giant waste of time. Especially the recursion assignment. That was pretty annoying.

  10. 10 lmm

    The first time we call mystery:
    t0 = [original return address]
    ra = [pc]
    jmp &here
    v0 = ra – 8 = &[last line of mystery] – 8 = &mystery (trusting the comment)
    return t0

    So the first call returns the address of mystery to C.
    I think technically assigning to A[2] and A[3] is illicit here, but meh. The point is you then fill in the assembly and the function calls what you wrote. And it needs to do different things on different calls, and presumably using ram is cheating.

    So what we want to do is something like:
    A[0]=add $v0,$t0,$t0 // v0 = t1 + t1
    A[1]=addiu $v0, $v0, 1 // v0 = v0 + 1
    A[2]=subu $t0, $v0, [original $t0] // t1 = v0 – [original t0]

    first time: returns original t0*2 + 1, stores original t0 + 1 in t0
    second time: returns original t0 * 2 + 3, stores original t0+3 in t0
    third time: returns original t0*2 + 7, stores original t0 + 7 in t0

    Only remaining thing is where we get our “original t0” from. So we need to find two registers that have the same value when the setup finishes, so we can use one to add to and one to keep the original value. Got to go now, will finish it off this evening.

  11. 11 lmm

    Oh, right, A is the address of the function (i.e. the original value of $t0), so we can use $a0 wherever I put [original t0] in the last post. Duh.

    (p.s. I don’t actually know MIPS at all. But I’m pretty sure that’s right)

  12. 12 nop

    I don’t know MIPS either (did a quick google) but I’m sure $t0 isn’t initialized to $a0 at any point, so I can’t tell how you’re using $a0 as the original $t0.

    From what I understand, $t0 contains whatever gibberish happens to be in there, unless parameters are passed both to $a0 and $t0? Maybe someone with more MIPS knowledge can enlighten me.

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