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level9
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imul - integer multiplication.

   0x080485c5 <+85>:	mov    -0x10(%ebp),%eax

    eax = ebp_10;


   0x080485c8 <+88>:	imul   -0x10(%ebp),%eax

    imul is a multiplication instruction.
    eax = eax * ebp_10;

    So, it is ebp_10 * ebp*10

   0x080485cc <+92>:	cmp    -0x8(%ebp),%eax
   0x080485cf <+95>:	jle    0x80485f0 <number_generator+128>

    if(ebp_10 * ebp_10 > ebp_8)...


idiv - integer division operation.
    Use a special instruction (cltd) right before idivl (idiv - long, 32bit),
    and 'cltd' (or CDQ) is for asserting using %edx:%eax as a 64bit integer,
    %edx for the top 32 bit (bit 63 - 32),
    and %eax for the lower 32 bit (bit 31 - 0).


   0x080485f0 <+128>:	mov    -0x8(%ebp),%eax

    eax = ebp_8

   0x080485f3 <+131>:	cltd

    edx = 0, eax = ebp_8, as a 64bit integer of %edx:%eax = ebp_8

   0x080485f4 <+132>:	idivl  -0x10(%ebp)

    divide that value (edx:eax) using ebp_10.

    Quotient will be stored in eax, and remainder will be stored in edx.

   0x080485f7 <+135>:	cmp    $0x0,%edx

    Check if the remainder is 0.


    So, if the assembly uses 'eax' for the result of idivl,
    you can interpret that as '/' operation,
    e.g., ebp_8 / ebp_10.

    But in this case, the assembly code uses 'edx' for the result of idivl,
    so you should interpret this as '%' operation, taking the remainder of
    division, e.g., ebp_8 % ebp_10.