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\begin{document}

Name: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Directions: {\bf Work only on this sheet} (on both sides, if needed); do
not turn in any supplementary sheets of paper. There is actually plenty
of room for your answers, as long as you organize yourself BEFORE
starting writing.

{\bf 1.} (10) Fill in the blank:  In addition to using a stack for
subroutines, the Intel hardware also uses a stack for
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ arithmetic.

{\bf 2.} Look at the assembler output on p.113.

\begin{itemize}

\item [(a)] (15) Suppose the instruction {\bf jz done} will placed
between lines 29 and 30.  What would the 75F8 for {\bf jnz top} change
to?

\item [(b)] (15) Suppose after linking, it has been decided that the {\bf
.data} section will begin at 0x00052000.  Then what will change, if
anything, in lines 24-34, and what will be the new value there if there
is a change?

\end{itemize}

{\bf 3.} (60) The following code goes through an array that is initially
pointed to by EAX, and searches in the array for the value in EBX.  The
array is terminated by a 0.  The result will be placed into EDX---either
the index at which the value was found, or -1 if it was not found.

For example, if the array is (1,5,2,13,0) and the value to be searched
for is 13, then 3 will be placed into EDX.  A search for 5 will result
in a 1 in EDX.  If the value to be searched for is 88, then -1 will be
placed into EDX.  Fill in the blanks:

\begin{Verbatim}[fontsize=\relsize{-2}]
      movl %eax, %edi
top:  movl (%eax), %ecx
                                      # put 1 instruction here
      jz foundit
                                      # put 1 instruction here
      jz notthere
                                      # put 1 instruction here
      jmp top  
notthere:
                                      # put 1 instruction here
      jmp done
foundit:
      subl %edi, %eax
                                      # put 1 instruction here
      movl %eax, %edx
done: addl $0, %esi  # dummy instruction
\end{Verbatim}

{\bf Solutions:}

{\bf 1.} floating-point

{\bf 2.a} The inserted JZ has a 2-byte code like the others, so the JNZ
will move 2 bytes further down.  That will change its distance to {\bf top}
from -8 to -10, the latter being 0xf6.  So, the new code for JNZ will be
75F6.

{\bf 2.b} The B900000000 will change to B900200500.  

The 891D10000000 will also change.  It is clear from inspection of lines
18 and 33 that register-to-memory MOV instructions have the format
891DIMM8, where IMM8 denotes an 8-byte constant.  That constant,
originally, 10000000, will change to 00200500 as above, so the new
instruction will be 891D00200500.

{\bf 3.}

\begin{Verbatim}[fontsize=\relsize{-2}]
      movl %eax, %edi
top:  movl (%eax), %ecx
      cmpl %ecx, %ebx
      jz foundit
      cmpl $0, %ecx
      jz notthere
      addl $4, %eax  
      jmp top  
notthere:
      movl $-1, %edx
      jmp done
foundit:
      subl %edi, %eax
      shrl $2, %eax
      movl %eax, %edx
done: dec %esi
\end{Verbatim}

\end{document}