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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

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fopen, fprintf, fclose

• Use ‘pointer’ to predetermined structureFILE *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 terminatesfclose( F );

• One way to write to a file – very similar to printf();fprintf( F, “test\n” );

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File write example #1

#include <stdio.h>

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,

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File read example #2

Output (on the monitor):Reading from the file:10, 20, 30, 40, 99, 32, 1, 999, -22, 4423,

#include <stdio.h>

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’:

1020304099321999-224423

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The content of the file ‘num10.dat’:

1,10.23, 205, 30.331,222 , 33345,469 40.113,-993.333399,33.2

Output (on the monitor):Reading from the file:1,10.23,205,30.331,222,245,469,463,-993.333

#include <stdio.h>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

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malloc / free#include <stdlib.h>

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);

}

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#include <stdio.h>#include <stdlib.h>#include <math.h>

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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From excel to C#include <stdio.h>

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, 0.881: 192607, -19.3, -6.242: 192608, -13.04, -2.83889: 200007, -3.55, 1.72890: 200008, -5.91, -5.17891: 200009, -10.62, -7.07892: 200010, -5.03, 1.58893: 200011, -5.7251, 0.245714

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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

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Examples#include <iostream.h> /* This is the stream definition file */

void print_it(int data_value);

main()

{

int index,top; /* A normal C variable */

cout << "Input a decimal value --> ";

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";

}

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class#include <iostream.h>

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 desiredvoid vehicle::initialize(int wheels, float weight){ wheels = in_wheels; weight = in_weight;}

// get the number of wheels of this vehicleint vehicle::get_wheels(){ return wheels;}

// return the weight of this vehiclefloat vehicle::get_weight(){ return weight;}

// return the weight on each wheelfloat 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.

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inheritanceclass 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.

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Newton Raphson method

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#include <stdio.h>#include <math.h>

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=4.001X=2.24994, Func(X)=0.0622188, Der=4.50X=2.23611, Func(X)=0.000204919, Der=4.The solution to Func(X)=0 is: X=2.23607

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#include <math.h>

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];

}

}

}

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Quiz - Question #1You 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");

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*****

*******

*********

*******

*****

***

*

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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?).

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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.