CS360 Lab #5 -- Assembler

  • Jim Plank
  • CS360

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    This is a written lab --- no programming. The TAs will tell you how best to get your answers to them. In these labs, you assume the architecture and instructions given in the lecture notes. Do not worry about delay slots. (In other words, I do not want to see any noop instructions).

    Do not "optimize" the assembler. Give me the simple yet inefficient assembler that the compiler would return.

    Question 1

    Show the assembler that the following code compiles into: 
    int b(int j)
{
  int i;

  i = 10;
  while (i > 0) {
    j = j + i * i - 3;
    i--;
  }
  return j;
}

    Question 2

    Show the assembler that the following code compiles into. You may assume that the value of NULL is zero. 
    int f2(int *x)
{
  int j;

  if (x == NULL) return 48;

  j = *x * 2;
  return j;
}

    Question 3

    1. Show the assembler that the following code compiles into. 
      int b(int i, int j)
{
  if (j <= 0) return i+1;

  return a(i, j-1)+1;
}

int a(int i, int j)
{
  if (i <= 0) return j+1;
  return b(i-1, j)+b(i-1, j-1);
}

main(int argc, char **argv)
{
  int i;

  if (argc != 3) { exit(1); }
  i = a(atoi(argv[1]), atoi(argv[2]));
}
    2. What is the final value of i when this program is called in the following ways: 
      UNIX> a.out 0 0
UNIX> a.out 1 2
UNIX> a.out 2 1
    3. Show the state of the stack and registers the first six times a jsr statement is reached when the program is called as follows: 
      UNIX> a.out 2 2
      Assume that the state of the stack and registers when main is first called looks as follows.

      Note, I'm happy with arrows instead of numeric values. 

                    Stack                              registers
        |----------------|                     |-----------------------|
        |                |                     |                       | r0
        |                |                     |                       | r1
        |    .....       |                     |                       | r2
        |    unused      |                     |                       | r3
        |    unused      |                     |                       | r4
        |    unused      |           /-------- |                       | sp
        |    unused      | <------------------ |                       | fp
        |   old fp       |                     | main:                 | pc
        |   old pc       |                     |-----------------------|
        |   argc = 3     |                
        |   argv         |--\             
/------ |   argv[0]      |<-/             
| /---- |   argv[1]      |                
| | /-- |   argv[2]      |                
\-|-|-> |'a'|'.'|'o'|'u' |
  | |   |'t'| 0 | 0 | 0  |
  \-|-> |'2'| 0 | 0 | 0  |
    \-> |'2'| 0 | 0 | 0  |
        |--------------- |

    Question 4

    1. Show the assembler that the following code compiles into (assume that the compiler doesn't check that the call of a in b has the wrong number of arguments). 
      int a(int i, int j, int *k)
{
  if (i == 0) {
    *k = 15;
    return j
  } else {
    return j;
  }
}

int b(int i)
{
  int m;

  m = i+25;

  return a(i);
}

main(int argc, char **argv)
{
  int i;
  i = b(atoi(argv[1]));
}
    2. Show the state of the stack right before a returns for each of the following invocations of the program: 
      UNIX> a.out 0
UNIX> a.out 1
    3. In each of the invocations above, what is the final value of i?
    Fri Feb 22 10:27:04 EST 2002