Arrays ELEC 206 Computer Applications for Electrical Engineers Dr. Ron Hayne Outline  Arrays  Statistical Measurements  Functions Revisited 206_C6 2 Arrays  One-dimensional array   List of values Arranged in either a row or a column  Offsets (Subscripts) distinguish between elements in the array   Identifier holds address of first element Offsets always start at 0 206_C6 3 Definition and Initialization  Declaration  double x[4];  Initialization    double x[4] = {1.2, -2.4, 0.8, 6.1}; int t[100] = {0}; char v[] = {'a', 'e', 'i', 'o', 'u'};  Loops double g[21]; for (int k=0; k<=20; k++) { g[k] = k*0.5; } 206_C6 4 Data Files double time[10], motion[10]; ifstream sensor3("sensor3.dat"); ... if(!sensor3.fail()) { for (int k=0; k<=9; k++) { sensor3 >> time[k] >> motion[k]; } } 206_C6 5 Function Arguments  Passing array information to a function  Two parameters usually used Specific array  Number of elements in the array   Always call by reference  Address of the array 206_C6 6 /* /* /* This program reads values from a data file and calls a function to determine the maximum value with a function. #include <iostream> #include <fstream> #include <string> using namespace std; // Define constants and declare function prototypes. const int N=100; double maxval(double x[], int n); int main() { // Declare objects. int npts=0; double y[N]; string filename; ifstream lab; */ */ */ ... // Read a data value from the file. while (npts <= (N-1) && lab >> y[npts] ) { npts++; // Increment npts. } // Find and print the maximum value. cout << "Maximum value: " << maxval(y,npts) << endl; // Close file and exit program. lab.close(); } return 0; } /* /* This function returns the maximum value in the array x with n elements. double maxval(double x[], int n) { // Declare local objects. double max_x; // Determine maximum value in the array. max_x = x[0]; for (int k=1; k<=n-1; k++) { if (x[k] > max_x) max_x = x[k]; } // Return maximum value. / return max_x; } */ */ Statistical Measurements      Maximum Minimum Mean (average) Median (middle) Variance  Average squared deviation from the mean  Standard Deviation  Square root of the variance 206_C6 10 Mean double mean(double x[], int n) { double sum(0); for (int k=0; k<=n-1; k++) { sum += x[k]; } return sum/n; } 206_C6 11 Variance double variance(double x[], int n) { double sum(0), mu; mu = mean(x,n); for (int k=0; k<=n-1; k++) { sum += (x[k] - mu)*(x[k] - mu); } return sum/(n-1); } 206_C6 12 Function Overloading  Function name can have more than one definition      Each function definition has a unique function signature Each function call looks for a function prototype with a matching function signature double maxval(double x[], int n); int maxval(int x[], int n); char maxval(char x[], int n); 206_C6 13 Function Templates  Generic algorithm for a function that is not tied to a specific data type template <class Data_type> Data_type minval(Data_type x[], int n) { Data_type minx; ... 206_C6 14 Summary  Arrays  Statistical Measurements  Functions Revisited 206_C6 15 Problem Solving Applied  Speech Signal Analysis  Problem Statement   Compute the following stastical measurements for a speech utterance: average power, average magnitude, and number of zero crossings. Input/Output Description Average power Average magnitude Zero crossings Zero.dat 206_C6 16 Problem Solving Applied  Hand Example  test.dat 2.5 8.2 -1.1 -0.2 1.5 Average power = 15.398  Average magnitude = 2.7  Number of zero crossings = 2  206_C6 17 Problem Solving Applied  Algorithm Development  main  read speech signal from data file and determine number of points, n  compute statistical measurements using functions  ave_power(x, n)  set sum to zero  for k: add x[k]2 to sum  return sum/n 206_C6 18 Problem Solving Applied  Algorithm Development  ave_mag(x, n)  set sum to zero  for k: add |x[k]| to sum  return sum/n  crossings(x, n)  set count to zero  for k: if x[k] * x[k+1] < 0, increment count  return count 206_C6 19 /*---------------------------------------------------*/ /* This program computes a set of statistical */ /* measurements from a speech signal. */ #include <cstdlib> #include <iostream> #include <fstream> #include <string> #include <cmath> using namespace std; // Declare function prototypes and define constants. double ave_power(double x[], int n); double ave_magn(double x[], int n); int crossings(double x[], int n); const int MAXIMUM = 2500; int main() { // Declare objects. int npts=0; double speech[MAXIMUM]; string filename; ifstream file_1; // Prompt user for file name and open file. cout << "Enter filename "; cin >> filename; file_1.open(filename.c_str()); if( file_1.fail() ) { cout << "error opening file " << filename << endl; return 1; } // Read information from a data file. while (npts <= MAXIMUM-1 && file_1 >> speech[npts]) { npts++; } // Compute and print statistics. cout << "Statistics \n"; cout << "\taverage power: " << ave_power(speech,npts) << endl; cout << "\taverage magnitude: " << ave_magn(speech,npts) << endl; cout << "\tzero crossings: " << crossings(speech,npts) << endl; // Close file and exit program. file_1.close(); system("PAUSE"); return 0; } /*---------------------------------------------------*/ /* This function returns the average power */ /* of an array x with n elements. */ double ave_power(double x[], int n) { // Declare and initialize objects. double sum=0; // Determine average power. for (int k=0; k<=n-1; k++) { sum += x[k]*x[k]; } // Return average power. return sum/n; } /*---------------------------------------------------*/ /* This function returns the average magnitude */ /* of an array x with n values. */ double ave_magn(double x[], int n) { // Declare and initialize objects. double sum=0; // Determine average magnitude. for (int k=0; k<=n-1; k++) { sum += abs(x[k]); } // Return average magnitude. return sum/n; } /*---------------------------------------------------*/ /* This function returns a count of the number */ /* of zero crossings in an array x with n values. */ int crossings(double x[], int n) { // Declare and initialize objects. int count=0; // Determine number of zero crossings. for (int k=0; k<=n-2; k++) { if (x[k]*x[k+1] < 0) count++; } // Return number of zero crossings. return count; } /*---------------------------------------------------*/ Testing 206_C6 26 Testing 206_C6 27 Sorting Algorithms  Sorting   Arranging in ascending (descending) order Not one "best" algorithm   Depends on characteristics of data Selection Sort Find smallest value from current position to end  Swap smallest with current position  Move to next position  206_C6 28 #include <iostream> #include <cstdlib> using namespace std; void sort(double x[], int n); void display(double x[], int n); const int N = 6; int main() { double data[N] = {5.0, 3.0, 12.0, 8.0, 1.0, 9.0}; cout << "Initial data" << endl; display(data, N); sort(data, N); cout << "Sorted data" << endl; display(data, N); ... void sort(double x[], int n) // Selection sort { int m; // marker double hold; for (int k=0; k<=n-2; k++) { m = k; // Find smallest value for (int j=k+1; j<=n-1; j++) { if (x[j] < x[m]) m = j; } hold = x[m]; // Swap smallest with current x[m] = x[k]; x[k] = hold; cout << "After k=" << k << endl; display(x, n); } } Testing 206_C6 31 Summary       Arrays Statistical Measurements Functions Revisited Problem Solving Applied Selection Sort End of Chapter Summary    C++ Statements Style Notes Debugging Notes 206_C6 32