EE150 - Spring 1996

Midterm #2 (4/1/96)
Solutions

Part A

The output is:

c=4, s=6

Here is the reason why.

The for loop differs from most we have seen in that the test involves a different variable (s) than the one we are initializing and incrementing (c).

The table below shows what happens each iteration of the loop (after doing the c++).

                 s   c
Before loop:     0   1
After 1st time:  1   2
After 2nd time:  3   3
After 3rd time:  6   4

The problem.

Part B

The output is:

?
*
**
***
?

Here's the reason why.

The loop runs i from 0 to 4, stopping when i is 5. When i is 0, we are going to the default case of the switch, writing a ?. When it is 1, 2 or 3, we do the appropriate case, writing out that many stars. When it is 4, we hit the default case again, writing the ?.

The problem.

Part C

The output is:

0.00
3

Here is the reason why.

The expression x/y is 10/20, where 10 and 20 are integers, so it results in a 0, which is converted to a double and assigned to d.

The expression d/y is 7.5/2, which will use real division and give us 3.25. However, this 3.25 is assigned to an integer x, which causes the .25 to be truncated, so x is assigned a 3.

The problem.

Part D

The output is:

4
17 9 13 17

Here is the reason why.

Before tweak is called the entire array t contains 17's. Inside tweak, changing s is equivalent to changing t, so we are modifying t[1] and t[2] to 9 and 13. But when we change t in tweak, that's changing the local variable and not the s that is passed to it.

The problem.

Problem #2

Part A

The changes are (any 4 out of these 6):

• Move the declaration of table out of main so that it is global.
• Add a global declaration for x.
• Pass 100 as an asgument to setTable.
• Add a prototype for setTable to main.
• Add a declaration for i to setTable.
• Add a return 0; to the end of main.

The problem.

Part B

The changes are:

• There should be an == instead of an = in the if statement.
• The assignment i = 1 should be i = 0.
• It should set t[i] to newvalue, not i.
• The function should return void rather than int.

The problem.

Part C

The changes are (any 4 out of these 5):

• The loop test should be against a newline, not EOF.
• The c=getchar() in the loop test needs to be surrounded by parentheses.
• The && should be an ||.
• The function should return int not void.
• c should be declared as an int.

The problem.

Problem #3

The key here is to write a loop that runs through the table and for each element, divides the world into cases: that of a positive table element and that of a negative table element. For a positive element, you need to write a set of dots, the !, the number of +'s in the table element's value, and then dots to the end. This winds up being a set of calls to writeN. The negative case is similar.

void displayTable(int a[], int n, int dots, char pos, char neg)
{
  void writeN(int c, int n);
  int i;

  for (i = 0; i < n; i++)
  {
    if (a[i] > 0)
    {
      writeN('.', dots);
      writeN('!', 1);
      writeN('+', a[i]);
      writeN('.', dots - a[i]);
    }
    else
    {
      writeN('.', dots + a[i]);
      writeN('-', -a[i]);
      writeN('!', 1);
      writeN('.', dots);
    }
    putchar('\n');
  }
}

The problem.

Problem #4

The key to this solution is to start with a loop that reads each character in the input, one character at a time. You then can construct a very simple state machine where you are either in a quote or not in a quote and reading a quote causes a transition between the states.

#include <stdio.h>

#define INQUOTE     0
#define NOT_INQUOTE 1

main()
{
  int c;
  int state = NOT_INQUOTE;

  while ((c = getchar()) != EOF)
    if (state == NOT_INQUOTE)
    {
      if (c == '"')
        state = INQUOTE;
    }
    else /* INQUOTE */
    {
      if (c != '"')
        putchar(c);
      else
      { 
        putchar('\n');
        state = NOT_INQUOTE;
      }
    }
  return 0;
}

#include 

main()
{
  int c;

  while ((c = getchar()) != EOF)
  {
    if (c == '"')    
    {
      while ((c = getchar()) != '"' && c != EOF)
        putchar(c);
      putchar('\n');
    }
  }
  return 0;
}

The problem.

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