1. EECE.2160 ECE Application Programming
Instructors:
Dr. Michael Geiger & Peilong Li
Spring 2016
Lecture 24
Strings

2. ECE Application Programming: Lecture 23
Lecture outline
Announcements/reminders
Program 4, 5 grades done; regrade deadlines TBD
Program 7 due 4/8 (will preview on Monday)
Today’s lecture
Character arrays and strings
Return exams
2/16/2019
ECE Application Programming: Lecture 23

3. ECE Application Programming: Lecture 23
Review: strings
Represented as character arrays
Can be initialized using string constants
char hello[] = “Hello”;
Can access individual elements
hello[3] = ‘l’;
Can print directly or with formatting
Print directly: printf(hello);
Print w/formatting using %s: printf(“%s\n”, hello);
Must leave enough room for terminating ‘\0’
2/16/2019
ECE Application Programming: Lecture 23

4. String functions (cont.)
In <string.h> library:
Copying strings:
char *strcpy(char *dest,
const char *source);
char *strncpy(char *dest,
const char *source,
size_t num);
Return dest
Does not append ‘\0’ unless length of source < num
Comparing strings:
int strcmp(const char *s1, const char *s2);
int strncmp(const char *s1, const char *s2, size_t num);
Character-by-character comparison of character values
Returns 0 if s1 == s2, >0 if s1 > s2, <0 if s1 < s2
2/16/2019
ECE Application Programming: Lecture 23

5. String functions (cont.)
Find # of characters in a string
size_t strlen(const char *s1);
Returns # characters before ‘\0’
Not necessarily size of array
“Add” strings together—string concatenation
char *strcat(char *dest,
const char *source);
char *strncat(char *dest,
const char *source,
size_t num);
Returns dest
2/16/2019
ECE Application Programming: Lecture 23

6. ECE Application Programming: Lecture 23
Example: Strings
What does the following program print?
int main() {
char s1[15];
int n1;
char s2[10] = “.216”;
int n;
strncpy(s1, “16”, 15);
n1 = strlen(s1);
printf(“s1 = %s\n”, s1);
printf(“Length of s1 = %d\n\n”, n1);
printf(“%c\n\n”, s1[1]);
strncat(s1,s2,10); n1 = strlen(s1); printf(“s1 = %s\n”, s1); printf(“Length of s1 = %d\n\n”, n1); // Assume user inputs: ABC ABD printf(“Enter two strings:”); scanf(“%s%s”, s1, s2); n = strncmp(s1, s2, 15); if (n > 0) printf(“%s > %s\n”, s1, s2); else if (n < 0) printf(“%s < %s\n”, s1, s2); else printf(“%s == %s\n”, s1, s2); return 0; }
2/16/2019
ECE Application Programming: Lecture 23

7. ECE Application Programming: Lecture 23
Example solution
s1 = 16 Initial value of s1 Length of s1 = 2 6 s1[1] s1 = s1 after strncat() Length of s1 = 6 Enter two strings: ABC ABD ABC < ABD Result of strncmp()
2/16/2019
ECE Application Programming: Lecture 23

8. Example: Using string functions
Works with main program in PE
Assume input strings have max of 49 chars (+ ‘\0’)
Write a function to do each of the following:
int readStrings(char *s);
Repeatedly read strings from standard input until the input string matches s. Return the number of strings read.
void copyNull(char *s1, char *s2, int n);
Copy the first n characters of s2 into s1, and make sure that the new version of s1 terminates with a null character.
int fillString(char *s, int n);
Repeatedly read strings from standard input and concatenate them to s until there is no room in the string. Return the final length of the string.
For example, if s is a 6-character array already holding “abcd”:
User enters “e”—string is full; return 5
User enters “ef”—there’s not enough room; return 4
Assume s initially contains null terminated string (or is empty)
2/16/2019
ECE Application Programming: Lecture 23

9. ECE Application Programming: Lecture 23
Example solution
int readStrings(char *s) { char str[50]; // Assume max 50 chars int count = 0; do { scanf(“%s”, str); // NOTE: do not // need &str count++; } while (strcmp(str, s) != 0); return count; }
2/16/2019
ECE Application Programming: Lecture 23

10. Example solution (cont.)
void copyNull(char *s1, char *s2, int n) {
strncpy(s1, s2, n);
s1[n] = ‘\0’; }
2/16/2019
ECE Application Programming: Lecture 23

11. Example solution (cont.)
int fillString(char *s, int n) {
char input[50]; // Assume max
// 50 chars
int charsLeft; // Space remaining
// in s
do {
scanf(“%s”, input);
// Calculate # chars left in array if input
// string is added. Need to leave room for ‘\0’
charsLeft = n – (strlen(s) + 1) – strlen(input);
if (charsLeft > 0) // Enough space to add this string
strcat(s, input); // and continue
else { // Out of room
if (charsLeft == 0) // Can add input, but then out of room
strcat(s, input);
return strlen(s); }
} while (1);
2/16/2019
ECE Application Programming: Lecture 23

12. ECE Application Programming: Lecture 23
Final notes
Next time
Structures
Reminders:
Program 4, 5 grades done; regrade deadlines TBD
Program 7 due 4/8 (will preview on Monday)
2/16/2019
ECE Application Programming: Lecture 23