16.216 ECE Application Programming

1. 16.216ECE Application Programming Instructor: Dr. Michael Geiger Fall 2013 Lecture 20 Arrays and functions (continued)

2. Lecture outline • Announcements/reminders • Program 6 due 10/28 • Review • Two-dimensional arrays • Arrays and functions • Today’s lecture • Two-dimensional arrays and functions • Character arrays and strings ECE Application Programming: Lecture 20

3. Review: arrays & pointers • Arrays: groups of data with same type • x[10] has 10 elements, x[0] through x[9] • Can also define with initial values • e.g. double list[] = {1.2, 0.75, -3.233}; • Must be sure to access inside bounds • Array name is a pointer • Arrays are always passed by address to functions • Can use pointer to access array • Can use arithmetic to move pointer through array ECE Application Programming: Lecture 20

4. Review: 2D arrays • Declared similarly to 1D arrays • Example (see below): int x[3][4]; • Index elements similarly to 1-D arrays • Initialize:int y[3][4] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12} }; • Typically used with nested for loops • Can pass to functions—must specify # columns ECE Application Programming: Lecture 20

5. 2-D arrays and functions • When passing 2-D array to function, can omit first dimension (rows) but must list columns • Example: // Assume n = # of rows int f(intarr[][4], int n); int main() { int x[3][4]; f(x, 3); ... } ECE Application Programming: Lecture 20

6. Example: 2-D arrays and functions • Say we have a program that stores student exam scores in a 2-D array: • Each row represents an individual student • Each column represents one of the 3 exams • Write functions to: • Calculate the exam average for each student and store it in a 1-D array that is accessible in the main program • Assume all exams have equal weight • Calculate the average for each exam and store it in a 1-D array that is accessible in the main program • Each function takes the same arguments: • The 2-D array • The # of students in the class • The 1-D array that will be used to hold the averages ECE Application Programming: Lecture 20

7. Example solution void studentAvg(double grades[][3], intnStudents, double averages[]) { inti, j; // Row/column # /* Go through each row, sum all columns, and divide by 3 to get each student’s avg */ for (i = 0; i < nStudents; i++) { averages[i] = grades[i][0]; // Initialize sum for (j = 1; j < 3; j++) { averages[i] += grades[i][j]; } averages[i] /= 3; } } ECE Application Programming: Lecture 20

8. Example solution (cont.) void examAvg(double grades[][3], intnStudents, double averages[]) { inti, j; // Row/column # /* Go through each column, sum all rows, and divide by nStudents to get each exam avg */ for (j = 0; j < 3; j++) { averages[j] = grades[0][j]; // Initialize sum for (i = 1; i < nStudents; i++) { averages[j] += grades[i][j]; } averages[j] /= nStudents; } } ECE Application Programming: Lecture 20

9. Strings in C • Strings in C: null-terminated arrays of characters • “Hello”{‘H’, ‘e’, ‘l’, ‘l’, ‘o’, 0} • Null character = 0 = ‘\0’ • Can declare array to hold string • Need space to hold null: char hello[5] would be too small • Can use string constants to directly initialize char hello[] = “Hello”; • Equivalent to: char hello[6]; hello[0] = ‘H’; hello[1] = ‘e’; hello[2] = ‘l’; hello[3] = ‘l’; hello[4] = ‘o’; hello[5] = 0 --OR-- hello[5] = ‘\0’; ECE Application Programming: Lecture 7

10. Strings, output, and functions • Can pass string as array or pointer: char * • printf(), scanf() take char * as first argument • Given string char hello[] from previous slide: • Print directly: printf(hello); • Print w/formatting using %s: printf(“%s\n”, hello); • Print individual character: printf(“%c\n”, hello[1]); ECE Application Programming: Lecture 7

11. String functions • Things we’d like to do with strings: • Set one equal to another • Compare two strings • Find # characters in string • String may not fill array (“buffer”) allocated for it • “Add” two strings together • “abc” + “def” = “abcdef” ECE Application Programming: Lecture 7

12. 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: • intstrcmp(const char *s1, const char *s2); • intstrncmp(const char *s1, const char *s2, size_t num); • Character-by-character comparison of character values • Returns 0 if s1 == s2, 1 if s1 > s2, -1 if s1 < s2 ECE Application Programming: Lecture 7

13. String functions (cont.) • Find # of characters in a string • size_tstrlen(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 ECE Application Programming: Lecture 7

14. 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; } ECE Application Programming: Lecture 7

15. Example solution s1 = 16 Initial value of s1 Length of s1 = 2 6 s1[1] s1 = 16.216 s1 after strncat() Length of s1 = 6 Enter two strings: ABC ABD ABC < ABD Result of strncmp() ECE Application Programming: Lecture 7

16. Final notes • Next time • Finish discussion of character arrays and strings • Reminders: • Program 6 due 10/28 ECE Application Programming: Lecture 20