16.216 ECE Application Programming

1. 16.216ECE Application Programming Instructor: Dr. Michael Geiger Fall 2012 Lecture 19: Math library One dimensional arrays

2. Lecture outline • Announcements/reminders • Program 5 due 10/26 • Today’s class • Math library • One dimensional arrays ECE Application Programming: Lecture 19

3. ANSI library functions • Example libraries • <stdio.h>: already using frequently • <stdlib.h>: various standard operations—conversion, memory allocation, etc. • <ctype.h> <string.h>: functions for manipulating characters and strings • We’ll cover these in a later lecture • <math.h>: Built-in math functions ECE Application Programming: Lecture 19

4. <math.h> functions • Trigonometric functions • All arguments of type double • Single argument in radians: • cos(r), sin(r), tan(r), acos(r), asin(r), atan(r) • Two arguments (x,y coordinates): atan2(x,y) • Remember: # in degrees = (180/pi) * (# in radians) • Exponential functions • exp(x): ex • log(x): natural log of x • log10(x): base 10 log of x • Power functions • pow(x, y): xy • sqrt(x): square root of x • Absolute value of a double: fabs() ECE Application Programming: Lecture 19

5. Example: Using math functions • Write a function to: • Given the other two sides, calculate and return the length of the hypotenuse of a right triangle • Remember, c2 = a2 + b2 • Given a number in degrees, calculate and return its sine • Remember: • Sine function takes an argument in radians • (# in degrees) = (180 / π) * (# in radians) • π = 4.0 * arctan(1.0) ECE Application Programming: Lecture 19

6. Example solutions • Write a function to: given the other two sides, calculate and return the length of the hypotenuse of a right triangle double calcHyp(double a, double b) { return sqrt(a * a + b * b); } ECE Application Programming: Lecture 19

7. Example solutions (cont.) • Write a function to: given a number in degrees, calculate and return its sine double calcSin(double deg) { double rad; rad = deg * (4.0 * atan(1.0)) / 180; return sin(rad); } ECE Application Programming: Lecture 19

8. Review of scalar variables Variables (up to now) have: name type (int, float, double, char) address value N 28C4 (int) q 28C8 (float) r 28CC (float) 35 3.14 8.9 e.g. Name type address value N integer 28C4 35 q float 28C8 3.14 r float 28CC 8.9 ECE Application Programming: Lecture 19

9. Intro to Arrays Any single element of x may be used like any other scalar variable x[0] 45 3044 x[1] 55 3048 x[2] 25 304C x[3] 85 3050 x[4] 75 3054 printf("%d",x[3]); // prints 85 printf("%d",x[7]+x[1]); // prints 115 x[5] 65 3058 x[6] 100 305C x[7] 60 3060 ECE Application Programming: Lecture 19

10. Declaring Arrays Define an 8 element array: int x[8]; Elements numbered 0 to 7 Arrays in C always start with location 0 (zero based) The initial value of each array element is unknown (just like scalar variables) x[0] ? 3044 x[1] ? 3048 x[2] ? 304C x[3] ? 3050 x[4] ? 3054 x[5] ? 3058 x[6] ? 305C x[7] ? 3060 ECE Application Programming: Lecture 19

11. Declaring/defining Arrays double A[]={ 1.23, 3.14159, 2.718, 0.7071 }; A[0] 1.23 4430 You can also define the values to be held in the array and instruct the compiler to figure out how many elements are needed. Not putting a value within the [] tells the compiler to determine how many locations are needed. A[1] 3.14159 4438 A[2] 2.718 4440 A[3] 0.7071 4448 ECE Application Programming: Lecture 19

12. Working with Arrays (input) #include <stdio.h> void main(void){ int x[8]; int i; // get 8 values into x[] for (i=0; i<8; i++) { printf("Enter test %d:",i+1); scanf("%d",&x[i]); } } ECE Application Programming: Lecture 19

13. Working with Arrays (input) Sample run (user input underlined): Enter test 1:80Enter test 2:75Enter test 3:90Enter test 4:100Enter test 5:65Enter test 6:88Enter test 7:40Enter test 8:90 x[0] 80 x[1] 75 x[2] 90 x[3] 100 x[4] 65 x[5] 88 x[6] 40 x[7] 90 ECE Application Programming: Lecture 19

14. Pitfalls • What happens if we change previous code to: #include <stdio.h> void main(void){int x[8];inti; float sum, avg; // used later // get 8 values into x[] for (i=0; i<=8; i++) {printf("Enter test %d:",i+1);scanf("%d",&x[i]); } } ECE Application Programming: Lecture 19

15. Pitfalls (cont.) • Although x has 8 elements, x[8] is not one of those elements! • Compiler will not stop you from accessing elements outside the array • Must make sure you know the size of the array ECE Application Programming: Lecture 19

16. Example • What does the following program print? int main() { intarr[10]; inti; printf("First loop:\n"); for (i = 0; i < 10; i++) { arr[i] = i * 2; printf("arr[%d] = %d\n", i, arr[i]); } printf("\nSecond loop:\n"); for (i = 0; i < 9; i++) { arr[i] = arr[i] + arr[i + 1]; printf("arr[%d] = %d\n", i, arr[i]); } return 0; } ECE Application Programming: Lecture 19

17. Example solution First loop: arr[0] = 0 arr[1] = 2 arr[2] = 4 arr[3] = 6 arr[4] = 8 arr[5] = 10 arr[6] = 12 arr[7] = 14 arr[8] = 16 arr[9] = 18 Output continued: Second loop: arr[0] = 2 arr[1] = 6 arr[2] = 10 arr[3] = 14 arr[4] = 18 arr[5] = 22 arr[6] = 26 arr[7] = 30 arr[8] = 34 ECE Application Programming: Lecture 19

18. Next time • Arrays and functions • Two dimensional arrays • Reminders: • Program 5 due 10/26 ECE Application Programming: Lecture 19