This section talks about functions. This is probably one of the earliest description of functions that people regularly read. Whatever the case, it’s a pretty simple description, though it’s interesting to see how little information you actually had to declare in the original C environment. There was very little error checking done by the compiler – I can’t imagine how hard it was to find bugs sometimes!

Exercise 1.15
This was easy enough. Just made a function. I start at 32 degrees Fahrenheit and increment by 4 until I get to 212.

int convert(int temp);

return hot;
}

This chapter introduces arrays and how to use them to count the number of characters in each of the words of an input file. Pretty simple and easy ideas. The problems are challenging though.

Exercise 1.13
This exercise wants you to make a histogram of the number of letters in each word. I used the state variable again to determine whether or not I was in a word. If I came across a space and the state was IN, then I knew I had just finished a word and could increment the proper number. The part that took awhile to figure out was to include a category for all words with more than ten letters in them. Once I got the program up and running that assumed all words were ten letters or less, which wasn’t hard, I essentially had to add if-else statements everywhere to check and see if the letter count was higher than ten. More tedious than anything, and I missed a couple, meaning I had to go back and do lots of debugging.

int c, i, j, state, count;
int ndigit[MAXLENGTH + 1]; //track words with 10+ letters separately

for (i = 0; i < MAXLENGTH + 1; ++i) {
printf("%d-letter words: ",i+1);
for(j = 0; j < ndigit[i]; j++) {
printf("X");
}
printf("\n");
}//for

Exercise 1-14
This program was pretty simple. Since any character I type in on my American keyboard is going to be an ASCII character of some type, I made an array of 256 spaces, just incremented each space when its ASCII code was used, and then only printed out the histogram for characters that showed up in the sentence. Once we get outside of ASCII this wouldn't work, but for basic letters, numbers, and punctuation, it was fine.

int c, i, j, state, count;
int ndigit[MAXLENGTH]; //track words with 10+ letters separately

while((c = getchar()) != 'q') {
++ndigit[c];
}//while
printf("\n\n\n");
for(i = 0; i < MAXLENGTH; i++) {
if(ndigit[i] > 0) {
printf("ASCII %d: ", i);
for(j = 0; j < ndigit[i]; j++)
printf("X");
printf(" (%d)", ndigit[i]);
printf("\n");
}//if
}//for

This chapter introduces a short little code that counts the number of lines, words, and characters in an input. It also introduces the OR operator, printed like || and the else operator, which matches the if statement. Both very useful. The program is pretty straightforward and simple.

Exercise 1.11
I guess to test this program you’d want to see if it could handle extreme cases. You could attempt to use the program to open a file that has nothing in it to see if it crashes. You could try a file with so many characters that you overflow the int that you are using to track characters. You could try other whitespace characters, maybe a lot in a row.

Exercise. 1.12
This was pretty simple – I used a state variable to determine whether or not the cursor was inside a word. If the next character was white space, and the state was set to IN, then I had just come to the end of a word, and printed a newline. Otherwise, nothing happens (which allows multiple white space characters to be in a row and still work right). If the character is an actually letter or number, then the state is set to IN and the character is printed.

int c, state;

while ((c = getchar()) != 'q') {
if (c == ' ' || c == '\n' || c == '\t') {
if(state == IN) {
state = OUT;
putchar('\n');
}
}//if
else {
state = IN;
putchar(c);
}//elseif
}//while