Learn to Program with C_ Learn to Program using the Popular C Programming Language ( PDFDrive )

Chapter 8 ■ Arrays
The first while loop will read characters until it reaches F, since F is the first alphabetic
character in the data. The second loop will store
F in word[0]
i in word[1]
r in word[2]
s in word[3]
t in word[4]
Since n is incremented after each character is stored, the value of n at this stage is 5. When
the space after t is read, the while loop exits and \0 is stored in word[5], properly terminating the
string. The array word will look like this:
We can now use word with any of the standard string functions and can print it using %s, as in:
printf("%s", word);
%s will stop printing characters when it reaches \0.
The above code is not perfect – we used it mainly for illustrative purposes. Since word is of
size 10, we can store a maximum of 9 letters (plus \0) in it. If the next word is longer than 9 letters
(for example, serendipity), the code will attempt to access word[10], which does not exist,
giving an “array subscript” error.
As an exercise, consider how you would handle words that are longer than what you have
catered for. (Hint: check that n is valid before storing anything in word[n].)
To illustrate how we can work with individual characters in a string, we write a function,
numSpaces, to count and return the number of spaces in a string str:
int numSpaces(char str[]) {
int h = 0, spaces = 0;
while (str[h] != '\0') {
if (str[h] == ' ') spaces++;
h++;
}
return spaces;
} //end numSpaces
Consider the code:
char phrase[] = "How we live and how we die";
printf("Number of spaces is %d\n", numSpaces(phrase));
The first statement creates an array of just the right size to hold the characters of the string
plus \0. Since the phrase contains 26 characters (letters and spaces), the array phrase will be of
size 27, with phrase[0] containing H, phrase[25] containing e and phrase[26] containing \0.
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