Sample Midterm Answers

What are all possible outputs of the following code fragment?
```
void f(int a, int b)
{
		printf("%d %d\n", a, b);
}
 
void main(void)
{
		int i = 5;
		f(++i, ++i);
}
```
Answer: The key pointis that the function arguments can be evaluated in any order. So, the function can be called as f(6, 7) or f(7, 6). So, the two possible outputs are:
```
6 7
```
and
```
7 6
```
Given the definitions
```
		int numbs[10];
		int *ptr = numbs;
```
which of the following are equivalent, and why?
1. numbs[3]
2. numbs + 3
3. * ( numbs + 3 )
4. * ( ptr + 3 )
5. *ptr + 3
Answer: A refers to the third element of the array numbs. B refers to the address of the third element of the array. C refers to the quantity at the address of the third element of the array, which is the third element of the array. As ptr is assigned numb, D refers to the same thing as C. E is the value of the element stored at numb, plus 3. Hence a, c, and d are equivalent.

Write a recursive function to add the integers from a to b. You may assume that a ≤ b initially. Answer:

int add(int a, int b)
{
		/* base case: a == b */
		if (a == b)
			return(b);
		/* add lowest element to sum of rest */
		return(a + add(a+1, b));
}

Use the following code fragment to answer parts a, b, and c:
```
for(x = i = 0; i <= 100; i += 2, x += i);
```
1. In one short sentence , what does this for loop do?
  Answer: It stores in x the sum of the even numbers from 0 to 100 inclusive.
2. Is the following while loop equivalent? If not, how does its result differ?
```
		x = i = 0;
		while( i++ = 100)
			x += ++i;
```
  Answer: No. The loop has a syntax error, because i++ cannot be assigned to. If the expression in the while condition were i++ <= 100, though, then the program will compile. However, the loop places in x sums the even numbers from 0 to 102, because when i is 100, i++ <= 100 is true (remember, the value of i is used before the "++" operator increments i).
3. Does the following for loop do the same thing? If not, what does it do?
```
		for(x = i = 0; i <= 100; i++){
			if (!(i % 2))
				continue;
			x = x + i;
		}
```
  Answer: No. This sums the odd integers from 0 to 100 inclusive and stores the value in x.
What does the following function do?
```
		int x(char *s, char *t)
		{
			for( ; *s == *t; s++, t++)
				if (*s == '\0')
					return(0);
			return(*s - *t);
		}
```
Answer: It retuns 0 if the two argument strings are the same, and the difference between the first characters in which the argument strings differ otherwise. This is (essentially) the strcmp(3) function.
What does this function do?
```
char *x(char *s, char c)
{
		char *r = NULL;
 
		do{
			while(*s && *s != c) s++;
			if (*s) r = s; 
		} while(*s++);
		return(r);
}
```
Answer: It returns a pointer to the last occurrence in argument s of the character in argument c. If the character does not occur in that string, it returns the NULL pointer.

The following segment of code is supposed to print the number of times the routine a_again is called. Yet, regardless of the input, it prints 0. Why? How would you fix it?

void a_again(int acount)
{
		++acount;
}
 
void main(void)
[
		register int c;
		int counter = 0;
 
		while((c = getchar()) != EOF)
			if (c == 'a' || c == 'A')
				a_again(counter);
 
		printf("%d\n", counter);
		exit(0);
}

Answer: The problem is that acount is passed as a parameter to a_again. As C calls by value, not reference, the value of counter is not changed by a_again. One way to fix this is to make counter global, and not pass anything to a_again:

int counter = 0;
void a_again()
{
		++counter;
}
 
void main(void)
[
		register int c;
 
		while((c = getchar()) != EOF)
			if (c == 'a' || c == 'A')
				a_again();
 
		printf("%d\n", counter);
		exit(0);
}