Midterm Key
CS31 (Shinnerl) Winter 06

Part I.

d b c b d c e c d d b f f a c e d b c

Part II.

#20

Constructs a positive integer from an array of digits stored as characters.
(Converts a C-style string representation of an integer value to its 
"int" representation.)

#21

a. 2
b. 3
c. 3
d. 3.75
e. 2.0

#22

Modular design: the program is organized as a collection 
of small, loosely coupled functions
interacting through prescribed interfaces.  I.e., preconditions
and postconditions on function parameters are carefully enforced.
Comments, consistent naming conventions and indentation, and 
structured control flow are also very important.


Part III.

#23

a.

int scoreCount(int a[], int n, int s)
{
	int count = 0;
	for (int i = 0; i < n; i++)
	{
		if (s == a[i])
			count++;
	}
	return count;
}

b.

int mode(int a[], int n)
{
	int index = 0;
	int count = 0;
	int highest = 0;
	for (int i = 0; i < n; i++)
	{
		count = scoreCount(a, n, a[i]);
		if (count > highest)
		{
			highest = count;
			index = i;
		}
	}
	return a[index];
}

#24

a.

/*
Find a match phrase starting from the pos in text
Compare each character until there is mismatch or
there is no more characters in phrase or text
Parameters:
	phrase - phrase to match
	text - text to match against
	pos - starting position in text
Returns:
	0 - no match
	1 - match
*/
int matchString(char phrase[], char text[], int pos)
{
	int i = 0;
	while (phrase[i] && text[pos])
	{
		if (phrase[i++] != text[pos++])
			return 0;
	}
	//reached the end of phrase?
	if (phrase[i])
		return 0;
	return 1;
}

int substringStart(char phrase[], char text[])
{
	int i = 0;
	while (text[i])
	{
		if (matchString(phrase, text, i))
			return i;
		i++;
	}
	return -1;
}


b.

int phraseCount(char phrase[], char text[])
{
	int count = 0;
	int i = 0;
	while (text[i])
	{
		if (matchString(phrase, text, i))
			count++;
		i++;
	}
	return count;
}