1. 1 C workshop #5 Files, Interface to Excel & Matlab, Introduction to C++, More Financial Applications (Newton-Raphson), Cholesky decomposision By: Yuli Kaplunovsky, yuli@magniel.com, (408) 309 4506

2. 2 fopen, fprintf, fclose Use ‘pointer’ to predetermined structure FILE *F; To create / open file: F = fopen( “file_name”, “type” ); examples: F = fopen(“file1.dat”, “w” ); F_Read = fopen(“file2.dat”,”r” ); ALWAYS need to close the file before program terminates fclose( F ); One way to write to a file – very similar to printf(); fprintf( F, “test\n” );

3. 3 File write example #1 #include void main() { FILE *F; int I; F = fopen("file1.dat", "w" ); fprintf( F, "ABCDE\n" ); fprintf( F, "Second Line\n" ); for ( I = 0 ; I < 10 ; I++ ) fprintf( F, "%d, ", I); fclose(F); } A new file by the name ‘file1.dat’ is created, and its content is: ABCDE Second Line 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,

4. 4 File read example #2 Output (on the monitor): Reading from the file: 10, 20, 30, 40, 99, 32, 1, 999, -22, 4423, #include void main() { FILE *F; int I, J; F = fopen("num10.dat", "rt" ); printf("Reading from the file:\n"); for ( I = 0 ; I < 10 ; I++ ) { fscanf( F, "%d", &J ); printf( "%d, ", J ); } fclose(F); } The content of the file ‘num10.dat’: 10 20 30 40 99 32 1 999 -22 4423

5. 5 The content of the file ‘num10.dat’: 1,10.2 3, 20 5, 30.33 1,2 22, 333 45,46 9 40.11 3,-993.3333 99,33.2 Output (on the monitor): Reading from the file: 1,10.2 3,20 5,30.33 1,2 22,2 45,46 9,46 3,-993.333 #include void main() { FILE *F; int I, J; char ST[200]; float FL; double D; F = fopen("num20.dat", "rt" ); printf("Reading from the file:\n"); for ( I = 0 ; I < 8 ; I++ ) { fgets( ST, sizeof(ST)-1, F ); sscanf( ST, "%d,%f", &J, &FL ); D = FL; printf( "%d,%g \n", J, D ); } fclose(F); } File read example #3

6. 6 malloc / free #include void main() { double *D; int I; D = (double *)malloc( sizeof(D) * 100 ); for ( I = 0 ; I < 100 ; I++ ) D[I] = (double)I * 1.2; free(D); }

7. 7 #include typedef struct { double X,Y; } CORD_typedef; int Generate_Sin_Noise( int N, char *FileName ) { FILE *F; int I; CORD_typedef *C; F = fopen(FileName, "w" ); if ( F == NULL ) return -1; // An Error C = (CORD_typedef *) malloc( sizeof(CORD_typedef) * N ); if ( C == NULL ) { fclose(F); return -1; // An Error } for ( I = 0 ; I < N ; I++ ) { C[I].X = (double)I / 30; C[I].Y = ((I==0)? 0 : C[I-1].Y) / 10.0 + (double)(rand() % 1001) / 100 + sin( C[I].X ) * 3.0; } for ( I = 0 ; I < N ; I++ ) fprintf( F, "%g,%g\n", C[I].X, C[I].Y ); fclose(F); delete(C); return 0; } void main() { if ( Generate_Sin_Noise( 200, "TestSin.dat" ) != 0 ) { printf("Error\n"); exit(-1); } printf("Done \n"); } Write file example

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14. 14 From excel to C #include void main() { FILE *F; char ST[300]; int ItemsNum,I; struct { int Month; double P1,P2; } Item[1000]; float F1,F2; F = fopen("Book1.txt", "rt" ); if ( F == NULL ) { printf("Error opening file\n"); return; } fgets(ST, sizeof(ST), F); // We know that the first line is not used ItemsNum = 0; while ( !feof(F) && ItemsNum < 1000 ) { fscanf( F, "%d %g %g\n", &(Item[ ItemsNum ].Month), &F1, &F2 ); Item[ ItemsNum ].P1 = F1; Item[ ItemsNum ].P2 = F2; ItemsNum++; } fclose(F); for ( I = 0 ; I < 3 ; I++ ) printf("%d: %d, %g, %g\n", I, Item[ I ].Month, Item[ I ].P1, Item[ I ].P2 ); for ( I = ItemsNum-5 ; I < ItemsNum ; I++ ) printf("%d: %d, %g, %g\n", I, Item[ I ].Month, Item[ I ].P1, Item[ I ].P2 ); } Output: 0: 192606, -12.31, 0.88 1: 192607, -19.3, -6.24 2: 192608, -13.04, -2.83 889: 200007, -3.55, 1.72 890: 200008, -5.91, -5.17 891: 200009, -10.62, -7.07 892: 200010, -5.03, 1.58 893: 200011, -5.7251, 0.245714

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17. 17 C++ object oriented programming Reusable code Abstract Data Types Encapsulate not necessary information MFC - Microsoft Foundation Classes But... Still it is almost the same C

18. 18 Examples #include /* This is the stream definition file */ void print_it(int data_value); main() { int index,top; /* A normal C variable */ cout "; cin >> top; for (index = 0 ; index < 5 ; index++) print_it(index); } void print_it(int data_value) { cout << "The value of the index is " << data_value << "\n"; }

19. 19 class #include class vehicle { protected: int wheels; float weight; public: void initialize(int wheels, float weight); int get_wheels(void); float get_weight(void); float wheel_loading(void); }; // initialize to any data desired void vehicle::initialize(int wheels, float weight) { wheels = in_wheels; weight = in_weight; } // get the number of wheels of this vehicle int vehicle::get_wheels() { return wheels; } // return the weight of this vehicle float vehicle::get_weight() { return weight; } // return the weight on each wheel float vehicle::wheel_loading() { return weight/wheels; } void main() { vehicle motorcycle, truck, sedan; motorcycle.initialize(2, 900.0); truck.initialize(18, 45000.0); sedan.initialize(4, 3000.0); cout << "The truck has a loading of " << truck.wheel_loading() << " pounds per wheel.\n"; cout << "The motorcycle weighs " << motorcycle.get_weight() << " pounds.\n"; cout << "The sedan weighs " << sedan.get_weight() << " pounds, and has " << sedan.get_wheels() << " wheels.\n"; } Output: The truck has a loading of 2500 pounds per wheel. The motorcycle weighs 900 pounds. The sedan weighs 3000 pounds, and has 4 wheels.

20. 20 inheritance class car : public vehicle { int passenger_load; public: void initialize(int in_wheels, float in_weight, int people = 4); int passengers(void); }; void car::initialize(int in_wheels, float in_weight, int people) { passenger_load = people; wheels = in_wheels; weight = in_weight; } int car::passengers(void) { return passenger_load; } void main() { car Mercedes; Mercedes.initialize(4, 3500.0, 5); cout << "The Mercedes carries " << Mercedes.passengers() << " passengers.\n"; cout << "The Mercedes weighs " << Mercedes.get_weight() << " pounds.\n"; cout << "The Mercedes's wheel loading is " << Mercedes.wheel_loading() << " pounds per tire.\n\n"; } Output: The Mercedes carries 5 passengers. The Mercedes weighs 3500 pounds. The Mercedes's wheel loading is 875 pounds per tire.

21. 21 Newton Raphson method

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23. 23 #include double Func( double X ) { return X*X-5; } double Newton_Raphson( double start, double accuracy ) { double X,Der; int Loop = 0; X = start; while ( fabs( Func(X) ) > accuracy ) { Der = (Func(X+0.001) - Func(X)) / 0.001; printf("X=%g, Func(X)=%g, Der=%g \n", X, Func(X), Der ); X = X - Func(X) / Der; if ( Loop++ > 20 ) { printf("There might be problem in finding a solution\n"); return X; } return X; } void main() { double D; D = Newton_Raphson( 2, 1e-6 ); printf("The solution to Func(X)=0 is: X=%g\n", D ); } Output: X=2, Func(X)=-1, Der=4.001 X=2.24994, Func(X)=0.0622188, Der=4.50 X=2.23611, Func(X)=0.000204919, Der=4. The solution to Func(X)=0 is: X=2.23607

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25. 25 #include void choldc(float **a, int n, float p[]) { void nrerror(char error_text[]); int i,j,k; float sum; for (i=1;i<=n;i++) { for (j=i;j<=n;j++) { for (sum=a[i][j],k=i-1;k>=1;k--) sum -= a[i][k]*a[j][k]; if (i == j) { if (sum <= 0.0) nrerror("choldc failed"); p[i]=sqrt(sum); } else a[j][i]=sum/p[i]; }

26. 26 Quiz - Question #1 You are asked to write a program that produces the following diamond shape. Try to minimize the number of printf functions. ONE PRINTF FUNCTION IS ALLOWED TO DISPLAY ONLY ONE CHARACTER, like printf(" "), or printf("*"), or printf("\n"), but not printf("*\n***\n*****\n"); * *** ***** ******* ********* ******* ***** *** *

27. 27 Question # 2 we want to count the number of primes between 2 and 100000. We will use an algorithm, which works as follows: Start with 2. Mark 4,6,8,10,...,100000, which are multiples of 2, as non-primes (you must find a way in C language to "mark" these numbers) Next is 3. Mark 6,9,12,15,..., 99999. Then 4, 5, 6,..., until 100000/2=50000. Those unmarked numbers must be primes (why?).

28. 28 Question # 3 Write a program to determine how many months it takes to pay off the purchase of a $1000 stereo system on credit. The purchase agreement was: no down payment, an interest rate of 18% per year (and hence 1.5% per month) and monthly payments of $50. The monthly payment is used to pay the interest due for that month and whatever is left after that is used to pay part of the remaining debt. Hence the first month you pay 1.5% of $1000 in interest - that is $15. So the remaining $35 is deducted from the debt, leaving a debt of $965. The next month you pay interest of 15% on the remaining $965, and etc. (The last payment may be less than $50.) Your program should determine the interest and principle paid each month, and the total interest paid over the course of the loan, and output that information along with the duration of the payments.