[SOLVED] CSCI4210-Operating Systems: Exam 1

Description

1. (6 POINTS) In the code below, what is output to STDOUT? Be careful. Circle the best answer.

int main() { char * s = “Operating systems rules”; s[10] = ‘S’; s[18] = ‘R’; printf(“%s\n”,s); }

• Operating systems rules
• <No Output to STDOUT>
• Operating systems rules\nOperating Systems Rules
• Operating Systems Rules
• Operating Systems Rules\nOperating systems rules

2. For a 64-bit architecture, how many total bytes are allocated on the heap and on the stack by the following code. Circle the best answer.

int main()

{ char * c[20]; char * s = calloc( 10, sizeof( int ) ); c[15] = s;

/* … */

}

• 80 stack, 80 heap (c) 0 stack, 120 heap      (e) 80 stack, 160 heap
• 160 stack, 40 heap (d) 200 stack, 0 heap

3. (6 POINTS) Given the code below, what is attached to the first five entries in the program file descriptor table? Circle the best answer.

int main()

{ close(1); int desc = open(“temp.txt”, O_CREAT|O_WRONLY|O_TRUNC); rc = dup2(desc, 3);

/* … */

}

• 0:STDIN, 1:<empty>, 2:STDERR, 3:”temp.txt”, 4:<empty>
• 0:STDIN, 1:”temp.txt”, 2:<empty>, 3:”temp.txt”, 4:<empty>
• 0:<empty>, 1:”temp.txt”, 2:STDERR, 3:”temp.txt”, 4:”temp.txt”
• 0:STDIN, 1:”temp.txt”, 2:STDERR, 3:”temp.txt”, 4:<empty>
• 0:”temp.txt”, 1:STDOUT, 2:”temp.txt”, 3:<empty>, 4:<empty>

4. (6 POINTS) Given the code below, how many processes are created? You should include the initial “parent” process in your count. You can assume all process identifiers begin at 100 and count up. Circle the best answer.

int main()

{ int ctr =0; while (ctr < 4) ctr += fork() + 1;

/* … */

}

• 2 (c) 5       (e) 6
• 11 (d) 14

5. (6 POINTS) Of the various non-preemptive CPU scheduling algorithms we have studied thus far, which scheduling algorithm is guaranteed to have the shortest average wait time? Circle the best answer.
   • FCFS (c) SRT (e) None of the above
   • SJF (d) RR
6. (6 POINTS) In the RR algorithm, the time slice is typically chosen … Circle the best answer.
   • as 20 milliseconds
   • to allow approximately 80% of CPU bursts to finish in one time slice.
   • to require approximately 50% of CPU bursts to take 2 or 3 time slices.
   • to require only the the longest 0.1% of CPU bursts to require more than one time slice.
   • as the clock speed of the processor times the number of processes in the wait queue.
7. (6 POINTS) What is the exact terminal output of the code below? Circle the best answer.

int main()

{

char * s = calloc( 100, sizeof( char ) ); strcpy( s, “RENSSELAER POLYTECHNIC”);

char * r = s + 20;

strcpy( r+3, “INSTITUTE”);

char * q = r – 5;

printf( “[%c][%c][%c][%s]\n”, *(s+9), *(q-4), *(r+3), s); return EXIT_SUCCESS;

}

• [R][P][I][RENSSELAER POLYTECHNIC INSTITUTE]
• [ ][P][N][RENSSELAER POLYTECHNIC INSTITUTE]
• [R][P][I][RENSSELAER POLYTECHNIC]
• [E][ ][ ][RENSSELAER POLYTECHNIC]
• None of the above

8. In SJT and SRT we estimate process running times based on exponential averaging. What is the formula for exponential averaging? Circle the best answer.
   • τ_i+1 = α∗t_i + (1 −α) ∗τ_i
   • None of a-d

For Questions 9 and 10, use the code shown below. As a hint, when fork() is called, any buffered output is copied from the parent process to the child process.

int main()

{ setvbuf( stdout, NULL, _IONBF, 0); close( 2 ); close( 1 ); int fd1 = open( “fd1.txt”, O_WRONLY | O_CREAT | O_TRUNC, 0644 ); int fd2 = open( “fd2.txt”, O_WRONLY | O_CREAT | O_TRUNC, 0644 ); fprintf( stderr, “01234” ); int rc = fork();

if ( rc > 0 )

{ wait( NULL ); printf( “56789” ); dup2(2,1);

} else { printf( “FGHIJ” ); fprintf( stderr, “ABCDE” );

} printf( “KLMNO” );

close(fd1); close(fd2); return EXIT_SUCCESS;

}

 

9. What is the exact contents of the txt file when this program completes its execution? Circle the best answer.
   • 01234ABCDEKLMNO
   • FGHIJKLMNO56789KLMNO (c) FGHIJKLMNO56789
   • FGHIJABCDEKLMNO56789KLMNO
   • 01234ABCDE
10. (6 POINTS) What is the exact contents of the txt file when this program completes its execution? Circle the best answer.
    • 01234ABCDEKLMNO
    • FGHIJKLMNO56789KLMNO (c) FGHIJKLMNO56789
    • FGHIJABCDEKLMNO56789KLMNO
    • 01234ABCDE
11. (6 POINTS) Given the code below, what is the contents of the txt file when the program terminates? Circle the best answer.

int main()

{ int fd = open( “q11.txt”, O_WRONLY | O_CREAT | O_TRUNC, 0766 ); write(fd, “First”, 5); close(fd); fd = open( “q11.txt”, O_WRONLY | O_CREAT , 0766 ); write(fd, “Next”, 3); close(fd);

/* … */

}

• FirstNext
• FirstNex
• Nex
• Nexst
• None of the above

12. The result of running this code would be? Circle the best answer.

int main()

{

int t = 19, u = 70; int * v = malloc( sizeof( int ) ); int * w = calloc( t, sizeof( int ) ); int * x = NULL;

*v = t;

free(w);

w = x; x = v;

printf( stderr, “%d %d\n”, *v, *w );

*x = 73; w = &u; fprintf( stderr, “%d %d\n”, *x, *w );

free( v ); return EXIT_SUCCESS;

}

• A buffer overflow
• A fork-bomb
• A segmentation fault
• A memory leak
• Buffered output

13. (16 POINTS) What is the exact terminal output of the code below? If multiple outputs are possible, succinctly describe all possibilities. Assume that the initial (parent) process ID is 200, with child processes, if any, numbered sequentially 300, 301, 302, etc.

int main()

{ int p[2]; int rc = pipe( p ); fprintf( stderr, “%d: %d %d\n”, getpid(), p[0], p[1] ); rc = fork();

if ( rc == 0 )

{ rc = write( p[1], “RPIirlRPIirlRPIirlRPI”, 12 );

♦	CPU Burst Time	Arrival Time
P1	12 ms	0 ms
P2	14 ms	2 ms
P3	4 ms	6 ms
P4	4 ms	13 ms

} else { rc = fork(); if (rc == 0)

{ rc = write( p[1], “OS-4210:OS-4210:OS-4210”, 12 );

} char buffer[13]; rc = read( p[0], buffer, 12); buffer[rc] = ‘\0’; fprintf(stderr, “%d: [%s]\n”, getpid(), buffer);

} return EXIT_SUCCESS;

}

14. (12 POINTS) Apply the SJF scheduling algorithm with the addition of aging to the process CPU bursts described in the table below. For each process, calculate the individual turnaround time, wait time, and number of preemptions experienced by each given process. Aging reduces the estimated running time of a process by 1ms for every 1ms spent in the wait queue. Aging is reset to 0 when a process returns to the wait queue and does not affect the actual burst time. Ignore context switch times and other such overhead.

Note that all “ties” are to be broken based on process ID order, lowest to highest.