Basics of Programming With Arrays

Outline • Review introduction to arrays with Basi Programm ing with Basics cs of Programmi Arrays – Declaring and using arrays – Array size and maximum subscript – Using a variable subscript for arrays • Writing code with arrays and for loops – Array sums – Finding maximum and minimum elements in an array • Data processing with arrays 2 Representing Data Run Data Math C++ 0 12.3 x0 x[0] 1 14.4 x1 x[1] 2 11.8 x2 x[2] 3 12.5 x3 x[3] 4 13.2 x4 x[4] 5 14.1 x5 x[5] Using Arrays • C++ array, x[i] used to represent data for which xi is used in mathematical notation • Declare arrays in typical way, but add maximum elements, e.g. int v[100]; • Refer to arrays as to any other variable using subscript v[3] or v[k] – Must assign value to k before using it as variable subscipt – Major tool in arrays is using variable subscript that is for loop index const int N = 200; double a[N]; for ( int j = 0; j < N; j++) a[j] = 0; 3 Maximum Array Subscript 4 General Array Processing • To process each element in an array with N elements, starting with the initial element, use a for loop with index k • Array subscripts start at zero • A declaration double y[N] declares a y array with N elements numbered from y[0] to y[N-1] N must be a const • For loop to handle all elements is – k starts at zero – The continuation condition, k < N, will process elements 0, 1, 2, … , N-1 – Increment k by 1 for ( int k = 0; k < N; k++ ) • C++ does not check to see if an array subscript is in bounds -- an incorrect subscript could affect some other memory location 5 • for ( int k = 0; k < N; k++) • Will process elements 0 to N – 1 of array regardless of array size 6 1 General Array Processing II General Array Processing III • On the previous chart N means the number of elements defined, not the total number of elements that can be stored in the array • Sometimes it is more convenient to refer to the subscripts than to the number of elements • E. g., array whose first and last defined elements have subscripts F and L • for ( k = F, k <= L; k++ ) • In the examples that follow, we will generally assume that an array has N elements, whose first subscript is zero • The for loop command to process each element in such an array is for ( k = 0; k < N; k++) • We can use different increments (e.g. k += 3) to skip elements 7 Setting Array to a Single Value 8 Keyboard Input, File Output const int MAX_SIZE = 100; double z[MAX_SIZE]; ofstream outFile( “array.out” ); for (int i = 0; i < MAX_SIZE; i++) { cout << “Enter z[“ << i << “]: “; cin >> z[i]; outFile << i << “ “ << z[i]; } • Simplest array processing code for (int k = 0; k < N; k++ ) arr[k] = value; • What is code to declare a double array, x, with 200 elements all set to zero? const int MAX = 200; double x[MAX] for (int k = 0; k < MAX; k++ ) x[k] = 0; 9 File Input, Screen Output 10 Computing Array Sums • Use previous tool of running sum variable along with for loop to include all array elements in the sum • Will use this code later to compute mean const int MAX_SIZE = 100; double z[MAX_SIZE]; ifstream inFile( “array.dat” ); for (int i = 0; i < MAX_SIZE; i++) { inFile >> z[i]; cout << “z[“ << i << “] = “ << z[i]; } int i; const int double x[N], sum = // Code to input N for( i = 0; i < N; { sum += x[i]; cout << “Array sum 11 N = …; 0; values i++ ) } = “ << of x[k] { } not needed here sum 12 2 Trace Array Sum Code k x[k] 0 12 1 14 2 11 3 12 4 13 5 14 Computing the Mean double x[N], x[6], sum = 0; for( i = 0; i < N; 6; i++ ) { sum += = sum x[i]; + x[i]; } Before loop sum = 0 i = 0: sum = 0 + 12 = 12 i = 1: sum = 12 + 14 = 26 i = 2: sum = 26 + 11 = 37 i = 3: sum = 37 + 12 = 49 i = 4: sum = 49 + 13 = 62 i = 5: sum = 62 + 14 = 76 i = 6: exit loop 1 x= N N 1 xi = ∑ N i =1 N −1 • N data items to average i =0 • Subscripts starts at 0 sum = 0; for( i = 0; i < N; i++ ) • Last data item { sum += x[i]; } is element N1 in array average = sum / N; • {} not needed ∑x i 13 14 Finding the Maximum Finding the Maximum • How do you find the maximum or minimum in a set of numbers? • E. g.: 13 74 -3 12 91 0 -17 88 -4 • Now that you found the maximum and minimum, how would you explain what you did so a computer can understand? • Scan the list and remember the largest (smallest) number you have seen and replace if you find one larger (smaller) • Store the current maximum in a variable (e.g., max) and compare max to new array values • If a new value is greater than max replace max by that value double max = x[0];// initialize max for ( int i = 1; i < N; i++ ) { if ( x[i] > max ) { max = x[i]; } } • Can omit both sets of braces 15 Finding the Minimum 16 Initializing Arrays • Store the current maximum in a variable (e.g., min) and compare min to new array values • If a new value is less than min replace min by that value double min = x[0];// initialize min for ( int i = 1; i < N; i++ ) { if ( x[i] < min ) { min = x[i]; } } • Can omit both sets of braces 17 • We can initialize an array by placing all the data values in braces following the array declaration int x[5] = { 12, 17, -22, 4, 12 }; const int NX = 5; int x[NX] = { 12, 17, -22, 4, 12 }; int x[] = { 12, 17, -22, 4, 12 }; • Note that the maximum size is not required when we initialize an array 18 3 Data Processing with Arrays Array Processing Example • In a typical experimental situation we can have fixed conditions (e.g. ambient pressure and temperature) – These are scalar (non-array) variables • We can have several data sets in which two or more variables are measured – Each variable measured in each run is an array variable – There is one array element for each run • You have taken current and voltage data from a circuit • There are N pairs of data • Current is stored as the amps[k] array and voltage as the volts[k] array • Write the code to compute the average power if arrays are already declared and data on N, volts[] and amps[] are already input 19 Average Power One 20 Average Power Two double sum = 0; for ( int k = 0; k < N; k++ ) { power[k] = amps[k] * volts[k]; sum += power[k]; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; double sum = 0 for ( int k = 0; k < N; k++ ) { power = amps[k] * volts[k]; sum += power; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; 21 22 Differences in Power Codes Average Power Three • Used three ways to compute power • Only one used a power[k] array • Code works with power not an array or not even a variable • Usually define arrays when we want to save results of a computation for use in subsequent computations double sum = 0 for ( int k = 0; k < N; k++ ) { sum += amps[k] * volts[k]; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; 23 24 4 Average Power One Average Power Two double sum = 0; for ( int k = 0; k < N; k++ ) { power[k] = amps[k] * volts[k]; sum += power[k]; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; The power array could be used in subsequent (or the same) loops double sum = 0 for ( int k = 0; k < N; k++ ) { power = amps[k] * volts[k]; sum += power; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; 25 26 Calculating Tables with Arrays Average Power Three • We have previously calculated tables where the step between variables was the same • How can we construct a table with uneven step sizes in the variables? double sum = 0 for ( int k = 0; k < N; k++ ) { sum += amps[k] * volts[k]; } double averagePower = sum / N; cout << “Power = “ << averagePower << “ watts”; The power calculation has to be redone if we want to use power[k] subsequently 27 Exercise – We can define one or two arrays for a oneway or two-way table – Each array has all the values that we want in the independent value of the table 28 Exercise II • The volume occupied by an idea gas is found from the equation V = nRT/P • Write a program that computes and prints V = nRT/P for a series of pressures using n = 1 kmol and T = 273.15 K – V is the volume in m3 – n is the number of kmols (kgmoles) – R = 8.3144 kPa· m3/kmol·K is the (universal) gas constant – T is the temperature in kelvins – P is the pressure in kPa • Write a program that computes V for fixed n and T and several P values The power variable can be used in the same loop only – P = 1, 2, 5, 10, 20, 50, 100, 200, 500, 1000, 2000, 5000, and 10000 kPa – R = 8.3144 kPa· m3/kmol·K 29 30 5 Exercise Solution const double R = 8.3144; const int NP = 13; double n = 1, T = 273.15, P[NP] = {1, 2, 5, 10, 20, 50, 100, 200, 500, 1000, 2000, 5000, 10000 }; for ( int k = 0; k < NP; k++ ) { cout << “\nFor P = “ << P[k] << “ kPa, V = “ << n * R * T / P[k] << “ m^3”; } 31 6