LAB 06 QUESTIONS SOLUTION

- Name: Tanner Hobbs
- NetID: Hobbs131

Answer the questions below according to the lab specification. Write
your answers directly in this text file and submit it to complete the
lab.


PROBLEM 1: ipow Assembly
========================

  The code pack contains the following files.
  - ipow_for.c : ipow function using a for loop
  - ipow_while.c : ipow function using a while loop
  - ipow_main.c : main function to call the above
  Both the for and while versions define the same function which has a
  meaning which should be obvious.

  Compile and run these as programs using either of the below:
  ,----
  | > gcc ipow_main.c ipow_for.c
  | > ./a.out 3 5
  | 3^5 = 243
  | 
  | OR
  | 
  | > gcc ipow_main.c ipow_while.c
  | > ./a.out 3 5
  | 3^5 = 243
  `----


A
~

  Compile ipow_for.c to assembly code. Look up how to do this with gcc
  if you did not take notes on how to do so from lecture. Make sure to
  disable optimizations by gcc when compiling.

  Paste the command you used and the generated assembly for the
  resulting ipow_for.s as you answer below.

    gcc -S ipow_main.c ipow_for.c

 Assembly code:

    .file    "ipow_for.c"
    .text
    .globl    ipow
    .type    ipow, @function
ipow:
.LFB0:
    .cfi_startproc
    pushq    %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    %edi, -20(%rbp)
    movl    %esi, -24(%rbp)
    movl    $1, -8(%rbp)
    movl    $0, -4(%rbp)
    jmp    .L2
.L3:
    movl    -8(%rbp), %eax
    imull    -20(%rbp), %eax
    movl    %eax, -8(%rbp)
    addl    $1, -4(%rbp)
.L2:
    movl    -4(%rbp), %eax
    cmpl    -24(%rbp), %eax
    jl    .L3
    movl    -8(%rbp), %eax
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size    ipow, .-ipow
    .ident    "GCC: (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0"
    .section    .note.GNU-stack,"",@progbits

B
~

  Similarly, compile the file ipow_while.c to assembly and compare the
  resulting ipow_while.s to ipow_for.s.
  - If you see differences, describe what these differences are and why
    the generated code varies between the two.
  - If you see similar code, describe why this might be.

   The code appears to be identical. This could be because while loops and for loops are essentially the same when boiled down.

C
~

  Examine the generated assembly code in ipow_for.s carefully. Based on
  the "assignments" of initial values to registers and the use of
  registers in operations like multiplication, addition, and comparison,
  attempt to draw a correspondence between the assembly registers and C
  variables. Ensure you identify the C variables
  - argument base: which register contains the 1st function argument?
  - argument exp: which register contains the 2nd function argument?
  - local pow which is also the return value
  - local i

  For your answer, paste in an annotated copy of the ipow_for.s which
  shows labels the correspondence of registers to C variables and
  describes how most of the assembly lines correspond to lines in the C
  code. You may skip lines like
  ,----
  | .cfi_startproc
  | .type    ipow, @function
  | .size    ipow, .-ipow
  | .ident    "GCC: (GNU) 7.3.0"
  | .section    .note.GNU-stack,"",@progbits
  `----
  which are assembler directives beyond the scope of our current
  interest.

      .file    "ipow_for.c"
    .text
    .globl    ipow
    .type    ipow, @function
ipow:
.LFB0:
    .cfi_startproc
    pushq    %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp           // function arguments 1 and 2 respectively
    .cfi_def_cfa_register 6
    movl    %edi, -20(%rbp)      // local pow variable
    movl    %esi, -24(%rbp)      // local i variable
    movl    $1, -8(%rbp)
    movl    $0, -4(%rbp)
    jmp    .L2
.L3:
    movl    -8(%rbp), %eax  
    imull    -20(%rbp), %eax
    movl    %eax, -8(%rbp)
    addl    $1, -4(%rbp)
.L2:
    movl    -4(%rbp), %eax
    cmpl    -24(%rbp), %eax
    jl    .L3
    movl    -8(%rbp), %eax
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size    ipow, .-ipow
    .ident    "GCC: (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0"
    .section    .note.GNU-stack,"",@progbits


D
~

  One oddity that is likely to appear in the generated assembly code is
  the instruction
  ,----
  | rep ret
  `----
  The `ret' instruction is a function return but `rep' is supposed to
  have a different purpose and the combination is strange.  Investigate
  this by
  - Reading page 208 of Bryant/O'Hallaron which describes this in an
    "aside" , OR
  - Doing a little internet search which should yield similar
    information.
  Describe roughly why `rep ret' shows up here.

rep shows up because of an error in amd's branch predictor.


PROBLEM 1: Coins Assembly code
==============================

  The code pack contains the following files.

  ------------------------------------------------------------------------
   FILE               Descriptions                                        
  ------------------------------------------------------------------------
   coins_main.c       main function to call the above                     
   coins_funcs.c      two functions manipulating the `coins_t' structure  
   coins_funcs_asm.s  incomplete assembly versions of the above functions 
  ------------------------------------------------------------------------

  The purpose of this problem is to complete the assembly code in
  `coins_funcs_asm.s' so that the functions there act identically to the
  C versions.

  Compile and the C version using the below:
  ,----
  | > gcc coins_main.c coins_funcs.c
  | > ./a.out 24
  | 24 cents is...
  | 0 quarters
  | 2 dimes
  | 0 nickels
  | 4 pennies
  | which is 24 cents
  | 
  | > ./a.out 63
  | 63 cents is...
  | 2 quarters
  | 1 dimes
  | 0 nickels
  | 3 pennies
  | which is 63 cents
  `----

  After completing the problems the following should compile and produce
  the same results.

  Compile and the C version using the below:
  ,----
  | > gcc coins_main.c coins_funcs_asm.s
  | > ./a.out 24
  | 24 cents is...
  | 0 quarters
  | 2 dimes
  | 0 nickels
  | 4 pennies
  | which is 24 cents
  `----

  Edit coins_funcs_asm.s to complete it and answer the questions below.


A
~

  Consider the first block of the assembly function marked BLOCK A.
  Describe what it is doing and how it relates to the C code in
  `coins_funcs.c'. Mention the purpose of the code around the label
  .OOB.


B
~

  BLOCK B performs a division operation. Based on the comments and
  assembly operations, describe what is going on, why the %eax register
  is used, and why the movb instruction is employed.
    movb 

C
~

  Complete BLOCK C by following the pattern laid out in the preceding
  blocks to calculate the number of nickels and pennies required and set
  the fields of the struct pointer to appropriate values. Consider
  carefully whether two additional divisions are required or if a single
  divide will suffice.


D
~

  In the total_coins function, study the pattern laid out the first few
  blocks which extract the 8-bit fields from the packed coins_t struct
  in register rdi. Note the use of shifts with AND operations to mask
  the low order bits to get a single field "alone" in a register before
  doing further operations. Copy this patter to complete BLOCK D which
  adds on nickels and pennies to the total.