Lesson of 23 July 2009

Use of the string class

File string-examples.cc

 1/*
 2  A simple program to test string class
 3 */
 4
 5#include <iostream> // for I/O
 6#include <iomanip>  // for hex, dec, width
 7#include <string>   // string support
 8
 9using namespace std ;
10
11int
12main() {
13  string a, b, c("$$$"), d("pippo") ;
14  
15  a = d ;
16  
17  char const * pa = a . c_str() ; // get the pointer to the C-string of a
18  char const * pd = d . c_str() ; // get the pointer to the C-string of d
19
20  cout << "address of string a = " << hex << setw(10) << (unsigned long)pa << '\n' ;
21  cout << "address of string d = " << hex << setw(10) << (unsigned long)pd << '\n' ;
22  //
23  //                                   ^       ^               ^
24  // use hexadecimal numer  -----------+       |               |
25  // set width for next output to 10 charater -+               |
26  // cast: convert type from pointer to unsigned integer ------+
27
28  a += " pluto" ;
29  cout << "a = '" << a << "'\n" ;
30  cout << "b = '" << b << "'\n" ;
31
32  pa = a . c_str() ; // get the pointer to the C-string of a
33  pd = d . c_str() ; // get the pointer to the C-string of d
34
35  cout << "address of string a = " << hex << setw(10) << (unsigned long)pa << dec << '\n' ;
36  cout << "address of string d = " << hex << setw(10) << (unsigned long)pd << dec << '\n' ;
37  
38  // main operation on string
39  
40  // concatenation, concatenate string a and c
41  b = a+c ;
42  cout << "a   = '" << a << "'\n" ;
43  cout << "c   = '" << c << "'\n" ;
44  cout << "a+c = '" << b << "'\n" ;
45  
46  // find
47  int i = a . find("plut") ;
48  cout << "a . find(\"plut\") = " << i << "\n" ;
49
50  i = a . find("asx") ;
51  cout << "a . find(\"asx\") = " << i << "\n" ;
52  
53  // comparison
54  a = "pippo" ;
55  b = "pluto" ; 
56  c = "paperino" ;
57  d = "pippo" ;
58  cout << "a = '" << a << "'\n" ;
59  cout << "b = '" << b << "'\n" ;
60  cout << "c = '" << c << "'\n" ;
61  cout << "d = '" << d << "'\n" ;
62  
63  cout << "'" << a <<"' < '" << b << "' = " ;
64  if ( a < b ) cout << "true\n" ;
65  else         cout << "false\n" ;
66
67  cout << "'" << a <<"' > '" << c << "' = " ;
68  if ( a > c ) cout << "true\n" ;
69  else         cout << "false\n" ;
70
71  cout << "'" << a <<"' == '" << d << "' = " ;
72  if ( a == d ) cout << "true\n" ;
73  else          cout << "false\n" ;
74
75  cout << "'" << b <<"' >= '" << c << "' = " ;
76  if ( b >= c ) cout << "true\n" ;
77  else          cout << "false\n" ;
78
79  cout << "'" << c <<"' >= '" << a << "' = " ;
80  if ( c >= a ) cout << "true\n" ;
81  else          cout << "false\n" ;
82  
83}

Use of the string class: convert a number to letter

File convert.cc

  1/*
  2  A simple program to convert a number to string.
  3  For example
  4  
  5  1234 => onethousandtwohundredthirtythree
  6
  7 */
  8
  9#include <iostream> // for I/O
 10#include <string>   // string support
 11
 12using namespace std ;
 13
 14string
 15convert0to19( unsigned n ) {
 16  // vector of strings for the first 20 numbers
 17  // *(str[]) = vector of pointer to character 
 18  char *str[] = {
 19    "", // 0 must be empty
 20    "one", // 1
 21    "two",
 22    "three",
 23    "four",
 24    "five",
 25    "six",
 26    "seven",
 27    "eight",
 28    "nine",
 29    "ten",
 30    "eleven",
 31    "twelve",
 32    "thirteen",
 33    "fourteen",
 34    "fifteen",
 35    "sixteen",
 36    "seventeen",
 37    "nineteen"
 38  } ;
 39  return str[ n % 20 ] ; // here the constructor string(str[n]) is used
 40                         // to build the return object
 41}
 42
 43string
 44convert0to99( unsigned n ) {
 45  // vector of strings for the first 20 numbers
 46  // *(str[]) = vector of pointer to character 
 47  char *str[] = {
 48    "",
 49    "ten",
 50    "twenty",
 51    "thirty",
 52    "fourty",
 53    "fifty",
 54    "sixty",
 55    "seventy",
 56    "eighty",
 57    "ninety",
 58  } ;
 59  
 60  if ( n < 20 ) return convert0to19(n) ;
 61  
 62  return string( str[ (n/10) % 10 ] ) + convert0to19( n % 10 )  ;
 63}
 64
 65string
 66convert0to999( unsigned n ) {
 67  // vector of strings for the first 20 numbers
 68  // *(str[]) = vector of pointer to character 
 69  char *str[] = {
 70    "",
 71    "onehundred",
 72    "twohundred",
 73    "threehundred",
 74    "fourhundred",
 75    "fivehundred",
 76    "sixhundred",
 77    "sevenhundred",
 78    "eighthundred",
 79    "ninehundred",
 80  } ;
 81
 82  if ( n < 100 ) return convert0to99(n) ;
 83  
 84  return string( str[ (n/100) % 10 ] ) + convert0to99( n % 10 )  ;
 85}
 86
 87string
 88convert0to999999( unsigned n ) {
 89  // vector of strings for the first 20 numbers
 90  // *(str[]) = vector of pointer to character 
 91
 92  if ( n < 1000 ) return convert0to999(n) ;
 93  
 94  unsigned upper = n / 1000 ;
 95  unsigned lower = n % 1000 ;
 96
 97  return convert0to999( upper ) + "thousand" + convert0to999( lower ) ;
 98}
 99
100string
101convert( unsigned n ) {
102  string res = convert0to999999( n ) ;
103  if ( res . length() > 0 ) return res ;
104  else                      return "zero" ; // call the contructor string("zero")
105}
106
107int
108main() {
109  cout << convert(10002)  << "\n" ;
110  cout << convert(21222)  << "\n" ;
111  cout << convert(1000)   << "\n" ;
112  cout << convert(991234) << "\n" ;
113}

Euler numerical scheme, C-like implementation

File Euler.cc

 1/*
 2  A simple example to use class and numerical integrator.
 3  Use Euler numerical method to "solve" a differential
 4  equation 
 5
 6  Y'   = f(t,Y)
 7  Y(0) = Y0
 8
 9 */
10
11#include <iostream> // for I/O
12#include <fstream> // for I/O
13#include <iomanip>  // for hex, dec, width
14#include <string>   // string support
15
16#include <algorithm>
17#include <vector>
18
19
20// define the problem
21//
22//  x' = y
23//  y' = -x 
24//  x(0) = 1
25//  y(0) = 0
26//
27//  Y = (x,y)^T  f(t,Y) = ( y, -x ) ^T
28//
29void
30f( double t, double Y[], double Y1[] ) {
31  Y1[0] = Y[1] ;
32  Y1[1] = -Y[0] ;
33}
34
35// Euler solver (C-way)
36void
37Euler( void (*fun)( double t, double Y[], double Y1[] ), // function to be integrated
38       double const Y0[], // initial value
39       int    const neq,  // number of equation
40       int    const N,    // number of steps
41       double const h,    // time step
42       double       Y[] ) { // output vector  
43
44  double * Fk = new double[neq] ;
45         
46  // copy the initial condition
47  std::copy( Y0, Y0 + neq, Y ) ;
48  for ( int k=0 ; k < N ; ++k ) {
49    double * Yk   = Y + neq * k ;
50    double * Ykp1 = Yk + neq ;
51    fun( k*h, Yk, Fk ) ;
52    // perform Euler step
53    for ( int i = 0 ; i < neq ; ++i ) 
54      Ykp1[i] = Yk[i] + h * Fk[i] ;
55  }
56
57  delete [] Fk ;
58}
59
60using namespace std ;
61
62int
63main() {
64  double Y[100][2] ;
65  double Y0[] = {1, 0} ; 
66  Euler( f, Y0, 2, 100, 0.1, (double*)Y ) ;
67  
68  ofstream file("out.txt") ;
69  file << "x\ty\n" ; // header of the output file
70  for ( int i=0 ; i < 100 ; ++i )
71    file << Y[i][0] << '\t' << Y[i][1] << '\n' ;
72  file . close() ;
73}

Euler numerical scheme, C++ implementation

File Euler1.cc

 1/*
 2  A simple example to use class and numerical integrator.
 3  Use Euler numerical method to "solve" a differential
 4  equation 
 5
 6  Y'   = f(t,Y)
 7  Y(0) = Y0
 8  
 9  better implementation using STL
10
11 */
12
13#include <iostream> // for I/O
14#include <fstream> // for I/O
15#include <iomanip>  // for hex, dec, width
16#include <string>   // string support
17
18#include <algorithm>
19#include <vector>
20
21using namespace std ;
22
23// define the problem
24//
25//  x' = y
26//  y' = -x 
27//  x(0) = 1
28//  y(0) = 0
29//
30//  Y = (x,y)^T  f(t,Y) = ( y, -x ) ^T
31//
32void
33f( double t, vector<double> const & Y, vector<double> & Y1 ) {
34  Y1[0] = Y[1] ;
35  Y1[1] = -Y[0] ;
36}
37
38// Euler solver
39template <typename T, int neq>
40void
41Euler( void (*fun)( T t, vector<T> const &, vector<T> &), // function to be integrated
42       vector<T> const Y0,   // initial value
43       int       const N,    // number of steps
44       T         const h,    // time step
45       vector<vector<T> > & Y ) { // output vector  
46
47  std::vector<T> Ykp1(neq), Fk(neq) ;
48  
49  Y . reserve( N ) ; // pre allocation for N elements
50
51  // copy the initial condition
52  std::copy( Y0 . begin(), Y0 . end(), Ykp1.begin() ) ;
53  Y . push_back( Ykp1 ) ;
54  
55  for ( int k=0 ; k < N ; ++k ) {
56    fun( k*h, Y.back(), Fk ) ;
57    // perform Euler step
58    for ( int i = 0 ; i < neq ; ++i ) 
59      Ykp1[i] = Y.back()[i] + h * Fk[i] ;
60    Y . push_back( Ykp1 ) ; // copy the last computed value to the end
61                            // of the output vector
62  }
63}
64
65int
66main() {
67  vector<vector<double> > Y ;
68  vector<double> Y0 ;
69  Y0 . push_back(1) ;
70  Y0 . push_back(0) ;
71
72  Euler<double,2>( f, Y0, 100, 0.1, Y ) ;
73
74  ofstream file("out.txt") ;
75  file << "x\ty\n" ; // header of the output file
76  
77  for ( vector<vector<double> >::iterator i = Y.begin() ;
78        i != Y . end() ; ++i )
79    file << (*i)[0] << '\t' << (*i)[1] << '\n' ;
80  file . close() ;
81}