Chapter 10 - Character Strings

1. Chapter 10 - Character Strings

2. Array of Characters char word[] = { ‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘!’ };

3. Program 10.1 /* * Function to concatenate two character strings */ #include <stdio.h> concat (char result[], char str1[], int n1, char str2[], int n2) { int i; // copy str1 to result for (i = 0; i < n1; i++) result[i] = str1[i]; for (i = 0; i < n2; i++) result[n1 + i] = str2[i]; }

4. Program 10.1 (continued) main() { char s1[5] = {'T', 'e', 's', 't', ' '}; char s2[6] = {'w', 'o', 'r', 'k', 's', '.'}; char s3[11]; int i; concat (s3, s1, 5, s2, 6); for (i = 0; i < 11; i++) printf("%c", s3[i]); printf ("\n"); system ("PAUSE"); }

5. Program 10.1 Output

6. Variable Length Character Strings char word[] = { ‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘!’, ‘\0’ }; NULL Character Terminates a String

7. Program 10.2 /* * Function to count the number of characters in a string */ #include <stdio.h> int string_length (char string[]) { int count = 0; while(string[count] != '\0') count++; return (count); }

8. Program 10.2 (continued) main() { char word1[] = {'a', 's', 't', 'e', 'r', '\0'}; char word2[] = {'a', 't', '\0'}; char word3[] = {'a', 'w', 'e', '\0'}; int i; printf ("%i %i %i\n", string_length (word1), string_length (word2), string_length (word3)); system ("PAUSE"); }

9. Program 10.2 Output

10. Initializing and Displaying Character Strings These Statements Are Equivalent char word[] = “Hello!”; char word[] = { ‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘!’, ‘\0’ }; char word[7] = “Hello!”; char word[6] = “Hello!”; Leave Room for the NULL Character

11. Program 10.3 /* * Function to concatenate two character strings */ #include <stdio.h> concat (char result[], char str1[], char str2[]) { int i; int j; // copy str1 to result for (i = 0; str1[i] != '\0'; i++) result[i] = str1[i]; for (j = 0; str2[j] != '\0'; j++) result[i + j] = str2[j]; result[i+j] = '\0'; }

12. Program 10.3 (continued) main() { char s1[] = "Test "; char s2[] = "works."; char s3[20]; concat (s3, s1, s2); printf("%s\n", s3); system ("PAUSE"); }

13. Program 10.3 Output

14. Testing Two Character Strings for Equality Since the C Programming Language does not support a data type of string we cannot directly test two strings to see if they are equal with a statement such as if ( string1 == string2 )

15. <string.h>

16. Program 10.4 /* * Function to determine if two strings are equal */ #include <stdio.h> #define TRUE 1 #define FALSE 0 int equal_strings (char s1[], char s2[]) { int i = 0; while ((s1[i] == s2[i]) && (s1[i] != '\0') && (s2[i] != '\0')) i++; if ((s1[i] == '\0') && (s2[i] == '\0')) return(TRUE); else return(FALSE); }

17. Program 10.4 (continued) main() { char stra[] = "string compare test"; char strb[] = "string"; printf ("%i\n", equal_strings (stra, strb)); printf ("%i\n", equal_strings (stra, stra)); printf ("%i\n", equal_strings (strb, "string")); system ("PAUSE"); }

18. Program 10.4 Output

19. Inputting Character Strings char string[81]; scanf ( “%s”, string); char s1[81], s2[81], s3[81]; scanf ( “%s%s%s”, s1, s2, s3);

20. Program 10.5 /* * Program to illustrate the %s scanf format characters */ #include <stdio.h> main() { char s1[81]; char s2[81]; char s3[81]; printf ("Enter text:\n"); scanf ("%s%s%s", s1, s2, s3); printf ("\ns1 = %s\ns2 = %s\ns3 = %s\n", s1, s2, s3); system ("PAUSE"); }

21. Program 10.5 Output

22. Program 10.6 /* * Function to read a line of text from a terminal */ #include <stdio.h> read_line (char buffer[]) { char character; int i = 0; do { character = getchar(); buffer[i] = character; i++; } while (character != '\n'); buffer[i-1] = '\0'; }

23. Program 10.6 (continued) main() { char line[81]; int i; for (i=0; i < 3; i++) { read_line(line); printf ("%s\n\n", line); } system ("PAUSE"); }

24. Program 10.6 Output

25. Program 10.7 #include <stdio.h> #define TRUE 1 #define FALSE 0 /* * Function to determine if a character is alphabetic */ alphabetic (char c) { if (('a' <= c && c <= 'z') || ('A' <= c && c <= 'Z')) return (TRUE); else return (FALSE); }

26. Program 10.7 (continued) /* * Function to count the number of words in a string */ count_words (char string[]) { int looking_for_word = TRUE; int word_count = 0; int i; for (i=0; string[i] != '\0'; i++) { if (alphabetic(string[i])) { if (looking_for_word) { word_count++; looking_for_word = FALSE; } } else looking_for_word = TRUE; } return (word_count); }

27. Program 10.7 (continued) main() { char text1[] = "Well, here goes."; char text2[] = "And here we go... again."; printf ("%s - words = %i\n", text1, count_words (text1)); printf ("%s - words = %i\n", text2, count_words (text2)); system ("PAUSE"); }

28. Program 10.7 Output

29. Execution of count_words Function

30. The NULL String A Character String that contains no characters other than the NULL Character has a special name in the C Programming Language, it is called the NULL String. The string length will correctly return 0. char buffer[100] = “”;

31. Program 10.8 #include <stdio.h> #define TRUE 1 #define FALSE 0 /***** Insert alphabetic function here *****/ /***** Insert read_line function here *****/ /***** Insert count_words function here *****/

32. Program 10.8 (continued) main() { char text[81]; int end_of_text = 0; int total_words = 0; printf ("Type in your text.\n"); printf ("When you are done, press 'RETURN'.\n\n"); while (!end_of_text) { read_line (text); if (text[0] == '\0') end_of_text = TRUE; else total_words += count_words (text); } printf("\nThere are %i words in the above text.\n", total_words); system ("PAUSE"); }

33. Program 10.8 Output

34. Escape Characters

35. More on Constant Strings If you put a backslash character at the very end of the line and followed it immediately by a carriage return, it will tell the C Compiler to ignore the end of line. This line continuation technique is used primarily for continuing long constant character strings. char letters[] = “abcdefghijklmnopqrstuvwxyz\ ABCDEFGHIJKLMNOPQRSTUVWXYZ”; An even easier way of breaking up long character strings is to divide them into two or more adjacent strings. char letters[] = “abcdefghijklmnopqrstuvwxyz” “ABCDEFGHIJKLMNOPQRSTUVWXYZ”;

36. Character Strings, Structures, and Arrays Suppose we wanted to write a computer program that acted as a dictionary. One of the first thoughts would be to representation of the word and it definition. Since the word and its definition are logically related, the notion of a structure comes immediately to mind. struct entry { char word[10]; char definition[50]; }; struct entry dictionary[100];

37. Program 10.9 /* * Dictionary lookup program */ #include <stdio.h> #define TRUE 1 #define FALSE 0 struct entry { char word[10]; char definition[50]; }; /***** Insert equal_string function here *****/

38. Program 10.9 /* * Function to lookup a word inside a dictionary */ int lookup (struct entry dictionary[], char search [], int entries) { int i; for (i=0; i < entries; i++) if (equal_strings (search, dictionary[i].word)) return (i); return (-1); }

39. Program 10.9 main() { struct entry dictionary[100] = { { "aardvark", "a burrowing African mammal" }, { "abyss", "a bottomless pit" }, { "acumen", "mentally sharp; keen" }, { "addle", "to become confused" }, { "aerie", "a high nest" }, { "affix", "to append; attach" }, { "agar", "a jelly made from seaweed" }, { "ahoy", "a nautical call of greeting" }, { "aigrette", "an ornamental cluster of feathers" }, { "ajar", "partially opened" } }; char word[10]; int entries = 10; int entry_number;

40. Program 10.9 printf("Enter word: "); scanf("%9s", word); entry_number = lookup (dictionary, word, entries); if (entry_number != -1) printf ("%s\n", dictionary[entry_number].definition); else printf ("Sorry, that word is not in my dictionary.\n"); system ("PAUSE"); }

41. Program 10.9 Output

42. A Better Search Method Binary Search Algorithm Step 1: Set low to 0, high to n – 1 Step2: If low > high, x does not exist in M and the algorithm terminates Step 3: Set mid to (low + high) / 2 Step 4: if M[mid] < x, set low to mid +1 and go to Step 2 Step 5: if M[mid] > x, set high to mid -1 and go to Step 2 Step 6: M[mid] equals x and the algorithm terminates

43. Binary Search Algorithm /* * Binary Search Algorithm */ int lookup (struct entry M[], char x[], int n) { int low = 0; /* Step 1: */ int high = n - 1; /* Step 1: */ int mid; int result; while (low <= high) { mid = (low + high) / 2; /* Step 3: */ result = compare_strings (M[mid].word, x); if (result == -1) low = mid + 1; /* Step 4: */ else if (result == 1) high = mid - 1; /* Step 5: */ else return (mid); /* Step 6: */ } return (-1); /* Step 2: */ }

44. Examples of Binary Search

45. Examples of Binary Search

46. Examples of Binary Search

47. Program 10.10 /* * Dictionary lookup program */ #include <stdio.h> struct entry { char word[10]; char definition[50]; };

48. Program 10.10 (continued) /* * Function to compare two character strings */ int compare_strings (char s1[], char s2[]) { int i = 0; while ((s1[i] == s2[i]) && (s1[i] != '\0') && (s2[i] != '\0')) i++; if (s1[i] < s2[i] ) /* s1 < s2 */ return(-1); else if (s1[i] == s2[i]) /* s1 == s2 */ return (0); else /* s1 > s2 */ return(+1); }

49. Program 10.10 (continued) /* * Function to lookup a word inside a dictionary */ int lookup (struct entry dictionary[], char search [], int entries) { int low = 0; int high = entries - 1; int mid; int result; while (low <= high) { mid = (low + high) / 2; result = compare_strings (dictionary[mid].word, search); if (result == -1) low = mid + 1; else if (result == 1) high = mid - 1; else return (mid); /* found it */ } return (-1); /* not found */ }

50. Program 10.10 (continued) main() { struct entry dictionary[100] = { { "aardvark", "a burrowing African mammal" }, { "abyss", "a bottomless pit" }, { "acumen", "mentally sharp; keen" }, { "addle", "to become confused" }, { "aerie", "a high nest" }, { "affix", "to append; attach" }, { "agar", "a jelly made from seaweed" }, { "ahoy", "a nautical call of greeting" }, { "aigrette", "an ornamental cluster of feathers" }, { "ajar", "partially opened" } }; char word[10]; int entries = 10; int entry_number;