veganafro / sampleexam-solutions

Sample Exam: Solutions

Problem 1: MCQs

1. b, d
2. b
3. b
4. a
5. d
6. c
7. a
8. a
9. b
10. b

Problem 2

Address (Before)	Value (Before)	Address (After)	Value (After)
0x400b028	0x00000022	0x400b028	0x00000032
0x400b024	0x400b611c	0x400b024	0x400b611c
0x400b020	0x400b512c	0x400b020	0x400b512c
0x400b01c	0x00000d12	0x400b01c	0x00000d12
0x400b018	0x00000014	0x400b018	0x00000014
0x400b014	0x400b511c	0x400b014	0x400b511c
0x400b010	0x400b601c	0x400b010	0x400b601c
0x400b00c	0x00000022	0x400b00c	Anything
0x400b008	0x00000013	0x400b008	Anything
0x400b004	0x400b601c	0x400b004	0x400b601c
0x400b000	0x400b511c	0x400b000	0x400b511c
0x400affc	0x00000013	0x400affc	0x00000032
0x400aff8	0x0000001B	0x400aff8	0x0000001B
0x400aff4	0x400b601c	0x400aff4	0x400b601c
0x400aff0	0x00000014	0x400aff0	0x00000014
0x400afec	0x0a0bc11d	0x400afec	0x0a0bc11d
0x400afe8	0x400b511d	0x400afe8	0x400b511d
0x400afe4	0x0000001B	0x400afe4	0x0000001B

Problem 3

A) The linker will generate an error because the strong symbol main is defined twice.

B) Output: x = 10

Problem 4

A) The code is not cache friendly. Because i is incremented in the inner loop, every successive access into the array is stride-n. We can make it cache-friendly by switching the loops:

int fun_1()
{
  int B[N][N];

  int i, j;

  for (i = 0 ; i < N ; i++)
    for (j = 0 ; j < N ; j++)
      B[i][j] = 2 * (B[i][j] + 2);
}

B) Going over the same array twice is not cache-friendly as we make no advantage of temporal locality. We can rewrite the function by using a single loop:

grade compute_average(grade* g, int n)
{
  float exam1_average = 0;
  float exam2_average = 0;
  int i;

  for (int i = 0 ; i < n ; i++)
  {
    exam1_average += g[i].exam1;
    exam2_average += g[i].exam2; 
  }

  grade average = { exam1_average / n, exam2_average / n};
  return average;
}

Problem 5

A) The actual struct in memory will look like:

struct node {
  char c;              /* 1 byte  */
  char pad1[7];        /* 7 bytes */
  long value;          /* 8 bytes */
  struct node* next;   /* 8 bytes */
  int flag;            /* 4 bytes */
  char pad2[4];        /* 4 bytes */
  struct node* left;   /* 8 bytes */
  struct node* right;  /* 8 bytes */
} node;

B) The blanks are:

• test1() --> movl 0x18(%rax), %eax
• test2() -->

	movq 0x10(%rax), %rax
	movq 0x8(%rax), %rax

Problem 6

This problem was discussed in recitation10 so you can refer to the solutions for that.

Problem 7

A) The output is the numbers: 0 1 2 1 2 2 2 (each on a separate line). A different ordering is also possible due to non-deterministic scheduling.

B) A total of 7 processes are created by fork(). This does not count main itself.

C) This is because main does not wait for the threads to finish. We can fix this by adding a join:

void main() { 
  pthread_t th[3];
  int i;
  for (i = 0; i < 3; i++) {
    pthread_create(&th[i], NULL, print_number, &i);
  }

  for (i = 0; i < 3; i++) {
    pthread_join(th[i], NULL);
  }

  exit(0); 
}

The output appears but is non-deterministic (i.e. changes every time) since we do not use synchronization. We are not required to do that.

D) There is only 1 process here since we create none.

E) main creates 3 threads.