C++ arrays and strstrok

I'm working with the arrays in C++. So my input in the array is O3B4F2 and I want to have in output OOOBBBBFF?.. I'm reading about function strtrok but I don't understand that really good because it's dividing sentence on the tokens.

#include <iostream>

#include <string.h>

#include<stdlib.h>

using namespace std;

int main ()

{

  char a[100+1];

  cin>>a;

  char * pch;

  char dioba="0 1 2 3 4 5  6 7 8 9 ";

  pch = strtok (a,dioba);

  int c;

  for(int i=0;i<strlen(a);i++)

  {

      if(isdigit(a[i])==1)

      {



      }

  }

  while (pch != NULL)

  {

    cout<<pch<<endl;

    pch = strtok (NULL,dioba);

  }



  return 0;

}

Also, I try to resolve a similar task where I need to divide the array into sets of the letter. I have output need togo and I want to my output look like ne e d to go. So after letter e an o, I want to use white space or new line.

 #include <iostream>

    #include <iostream>

    #include <string.h>

    #include<stdlib.h>

    using namespace std;

    int main ()

    {

      char a[100+1];

      cin>>a;

      char b[100+1];

      int  i=0,j=0;

      for(i;i<strlen(a);i++)

       {

        if(a[i]=='a'||a[i]=='e'|| a[i]=='i')

            for(j;j<strlen(a);j++)

            {

                b[j]=' ';

            }

          b[j]= a[i];

         cout<<b<<endl;

       }

      return 0;

     }

asked Nov 22 '18 at 7:52

Be Behind

197

You can use walk the input char by index or by pointer, no need to use strtok. As for mapping digit chars to numbers, a common trick is to use c - '0' to get the digit value.

– Botje
Nov 22 '18 at 8:09

Re: if(isdigit(a[i])==1) -- this won't work. Read the documentation for std::is digit.

– Pete Becker
Nov 22 '18 at 14:52

add a comment |

#include <iostream>

#include <string.h>

#include<stdlib.h>

using namespace std;

int main ()

{

  char a[100+1];

  cin>>a;

  char * pch;

  char dioba="0 1 2 3 4 5  6 7 8 9 ";

  pch = strtok (a,dioba);

  int c;

  for(int i=0;i<strlen(a);i++)

  {

      if(isdigit(a[i])==1)

      {



      }

  }

  while (pch != NULL)

  {

    cout<<pch<<endl;

    pch = strtok (NULL,dioba);

  }



  return 0;

}

 #include <iostream>

    #include <iostream>

    #include <string.h>

    #include<stdlib.h>

    using namespace std;

    int main ()

    {

      char a[100+1];

      cin>>a;

      char b[100+1];

      int  i=0,j=0;

      for(i;i<strlen(a);i++)

       {

        if(a[i]=='a'||a[i]=='e'|| a[i]=='i')

            for(j;j<strlen(a);j++)

            {

                b[j]=' ';

            }

          b[j]= a[i];

         cout<<b<<endl;

       }

      return 0;

     }

asked Nov 22 '18 at 7:52

Be Behind

197

You can use walk the input char by index or by pointer, no need to use strtok. As for mapping digit chars to numbers, a common trick is to use c - '0' to get the digit value.

– Botje
Nov 22 '18 at 8:09

Re: if(isdigit(a[i])==1) -- this won't work. Read the documentation for std::is digit.

– Pete Becker
Nov 22 '18 at 14:52

add a comment |

#include <iostream>

#include <string.h>

#include<stdlib.h>

using namespace std;

int main ()

{

  char a[100+1];

  cin>>a;

  char * pch;

  char dioba="0 1 2 3 4 5  6 7 8 9 ";

  pch = strtok (a,dioba);

  int c;

  for(int i=0;i<strlen(a);i++)

  {

      if(isdigit(a[i])==1)

      {



      }

  }

  while (pch != NULL)

  {

    cout<<pch<<endl;

    pch = strtok (NULL,dioba);

  }



  return 0;

}

 #include <iostream>

    #include <iostream>

    #include <string.h>

    #include<stdlib.h>

    using namespace std;

    int main ()

    {

      char a[100+1];

      cin>>a;

      char b[100+1];

      int  i=0,j=0;

      for(i;i<strlen(a);i++)

       {

        if(a[i]=='a'||a[i]=='e'|| a[i]=='i')

            for(j;j<strlen(a);j++)

            {

                b[j]=' ';

            }

          b[j]= a[i];

         cout<<b<<endl;

       }

      return 0;

     }

asked Nov 22 '18 at 7:52

Be Behind

197

#include <iostream>

#include <string.h>

#include<stdlib.h>

using namespace std;

int main ()

{

  char a[100+1];

  cin>>a;

  char * pch;

  char dioba="0 1 2 3 4 5  6 7 8 9 ";

  pch = strtok (a,dioba);

  int c;

  for(int i=0;i<strlen(a);i++)

  {

      if(isdigit(a[i])==1)

      {



      }

  }

  while (pch != NULL)

  {

    cout<<pch<<endl;

    pch = strtok (NULL,dioba);

  }



  return 0;

}

 #include <iostream>

    #include <iostream>

    #include <string.h>

    #include<stdlib.h>

    using namespace std;

    int main ()

    {

      char a[100+1];

      cin>>a;

      char b[100+1];

      int  i=0,j=0;

      for(i;i<strlen(a);i++)

       {

        if(a[i]=='a'||a[i]=='e'|| a[i]=='i')

            for(j;j<strlen(a);j++)

            {

                b[j]=' ';

            }

          b[j]= a[i];

         cout<<b<<endl;

       }

      return 0;

     }

c++ arrays strtok typed-arrays

asked Nov 22 '18 at 7:52

Be Behind

197

asked Nov 22 '18 at 7:52

Be Behind

197

asked Nov 22 '18 at 7:52

Be Behind

197

asked Nov 22 '18 at 7:52

Be Behind

197

asked Nov 22 '18 at 7:52

Be Behind

197

You can use walk the input char by index or by pointer, no need to use strtok. As for mapping digit chars to numbers, a common trick is to use c - '0' to get the digit value.

– Botje
Nov 22 '18 at 8:09

Re: if(isdigit(a[i])==1) -- this won't work. Read the documentation for std::is digit.

– Pete Becker
Nov 22 '18 at 14:52

add a comment |

You can use walk the input char by index or by pointer, no need to use strtok. As for mapping digit chars to numbers, a common trick is to use c - '0' to get the digit value.

– Botje
Nov 22 '18 at 8:09

Re: if(isdigit(a[i])==1) -- this won't work. Read the documentation for std::is digit.

– Pete Becker
Nov 22 '18 at 14:52

You can use walk the input char by index or by pointer, no need to use strtok. As for mapping digit chars to numbers, a common trick is to use c - '0' to get the digit value.

– Botje
Nov 22 '18 at 8:09

Re: if(isdigit(a[i])==1) -- this won't work. Read the documentation for std::is digit.

– Pete Becker
Nov 22 '18 at 14:52

add a comment |

2 Answers
2

active

oldest

votes

For your first case, as mentionned in comments, no need to use strtok. Here is an example of code (although there are many ways to perform the requested task):

#include <iostream>

#include <string>

#include <sstream>



int main ()

{

  std::string s;

  std::cin >> s;

  std::istringstream stream(s);



  char pch;



  // fetch stream for the character to repeat until the end of the string

  while( stream >> pch )

  {

      char nbChars;

      // fetch length for repetition

      stream >> nbChars;

      // convert character to its integer value

      nbChars -= '0';

      // repeat character as many times as needed

      for(int i=0; i < nbChars;i++)

      {

          std::cout << pch;

      }

  }



  return 0;

}

For your second task, I propose this sample code, you can modify letters as needed.

#include <iostream>

#include <string>



int main ()

{

  std::string s;

  std::string needSpaceChars{"aeo"};



  // read content on standard input

  std::getline(std::cin, s);

  // for each char, check its value and add space after the letters defined in needSpaceChars variable

  for(char pch: s)

  {

      // display character first

      std::cout << pch;

      // add space if character is in the list of characters to handle

      if( needSpaceChars.find(pch) != std::string::npos )

      {

          std::cout << ' ';

      }

  }



  return 0;

}

answered Nov 22 '18 at 9:01

Aurelien Normand

491

add a comment |

The way you have described the problem, the input will always in form of char followed by int. Hence you have to pick in pair (char, int)and print it as per rule.

for(size_t i = 0; i < arr.size() - 1;) {

    int val = (int)arr[i+1] - 48;

    for(auto j = 0; i < val; j++) {

        cout << arr[i];

    }

    i += 2;

}

First for() is for traversing through whole arr.

Second for() is for printing the particular char for it's number of occurrences.

Note : Since the given arr is of type char you have to convert the second element of pair from char to int.

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

For your first case, as mentionned in comments, no need to use strtok. Here is an example of code (although there are many ways to perform the requested task):

#include <iostream>

#include <string>

#include <sstream>



int main ()

{

  std::string s;

  std::cin >> s;

  std::istringstream stream(s);



  char pch;



  // fetch stream for the character to repeat until the end of the string

  while( stream >> pch )

  {

      char nbChars;

      // fetch length for repetition

      stream >> nbChars;

      // convert character to its integer value

      nbChars -= '0';

      // repeat character as many times as needed

      for(int i=0; i < nbChars;i++)

      {

          std::cout << pch;

      }

  }



  return 0;

}

For your second task, I propose this sample code, you can modify letters as needed.

#include <iostream>

#include <string>



int main ()

{

  std::string s;

  std::string needSpaceChars{"aeo"};



  // read content on standard input

  std::getline(std::cin, s);

  // for each char, check its value and add space after the letters defined in needSpaceChars variable

  for(char pch: s)

  {

      // display character first

      std::cout << pch;

      // add space if character is in the list of characters to handle

      if( needSpaceChars.find(pch) != std::string::npos )

      {

          std::cout << ' ';

      }

  }



  return 0;

}

answered Nov 22 '18 at 9:01

Aurelien Normand

491

add a comment |

For your first case, as mentionned in comments, no need to use strtok. Here is an example of code (although there are many ways to perform the requested task):

#include <iostream>

#include <string>

#include <sstream>



int main ()

{

  std::string s;

  std::cin >> s;

  std::istringstream stream(s);



  char pch;



  // fetch stream for the character to repeat until the end of the string

  while( stream >> pch )

  {

      char nbChars;

      // fetch length for repetition

      stream >> nbChars;

      // convert character to its integer value

      nbChars -= '0';

      // repeat character as many times as needed

      for(int i=0; i < nbChars;i++)

      {

          std::cout << pch;

      }

  }



  return 0;

}

For your second task, I propose this sample code, you can modify letters as needed.

#include <iostream>

#include <string>



int main ()

{

  std::string s;

  std::string needSpaceChars{"aeo"};



  // read content on standard input

  std::getline(std::cin, s);

  // for each char, check its value and add space after the letters defined in needSpaceChars variable

  for(char pch: s)

  {

      // display character first

      std::cout << pch;

      // add space if character is in the list of characters to handle

      if( needSpaceChars.find(pch) != std::string::npos )

      {

          std::cout << ' ';

      }

  }



  return 0;

}

answered Nov 22 '18 at 9:01

Aurelien Normand

491

add a comment |

For your first case, as mentionned in comments, no need to use strtok. Here is an example of code (although there are many ways to perform the requested task):

#include <iostream>

#include <string>

#include <sstream>



int main ()

{

  std::string s;

  std::cin >> s;

  std::istringstream stream(s);



  char pch;



  // fetch stream for the character to repeat until the end of the string

  while( stream >> pch )

  {

      char nbChars;

      // fetch length for repetition

      stream >> nbChars;

      // convert character to its integer value

      nbChars -= '0';

      // repeat character as many times as needed

      for(int i=0; i < nbChars;i++)

      {

          std::cout << pch;

      }

  }



  return 0;

}

For your second task, I propose this sample code, you can modify letters as needed.

#include <iostream>

#include <string>



int main ()

{

  std::string s;

  std::string needSpaceChars{"aeo"};



  // read content on standard input

  std::getline(std::cin, s);

  // for each char, check its value and add space after the letters defined in needSpaceChars variable

  for(char pch: s)

  {

      // display character first

      std::cout << pch;

      // add space if character is in the list of characters to handle

      if( needSpaceChars.find(pch) != std::string::npos )

      {

          std::cout << ' ';

      }

  }



  return 0;

}

answered Nov 22 '18 at 9:01

Aurelien Normand

491

For your first case, as mentionned in comments, no need to use strtok. Here is an example of code (although there are many ways to perform the requested task):

#include <iostream>

#include <string>

#include <sstream>



int main ()

{

  std::string s;

  std::cin >> s;

  std::istringstream stream(s);



  char pch;



  // fetch stream for the character to repeat until the end of the string

  while( stream >> pch )

  {

      char nbChars;

      // fetch length for repetition

      stream >> nbChars;

      // convert character to its integer value

      nbChars -= '0';

      // repeat character as many times as needed

      for(int i=0; i < nbChars;i++)

      {

          std::cout << pch;

      }

  }



  return 0;

}

For your second task, I propose this sample code, you can modify letters as needed.

#include <iostream>

#include <string>



int main ()

{

  std::string s;

  std::string needSpaceChars{"aeo"};



  // read content on standard input

  std::getline(std::cin, s);

  // for each char, check its value and add space after the letters defined in needSpaceChars variable

  for(char pch: s)

  {

      // display character first

      std::cout << pch;

      // add space if character is in the list of characters to handle

      if( needSpaceChars.find(pch) != std::string::npos )

      {

          std::cout << ' ';

      }

  }



  return 0;

}

answered Nov 22 '18 at 9:01

Aurelien Normand

491

answered Nov 22 '18 at 9:01

Aurelien Normand

491

answered Nov 22 '18 at 9:01

Aurelien Normand

491

answered Nov 22 '18 at 9:01

Aurelien Normand

491

add a comment |

The way you have described the problem, the input will always in form of char followed by int. Hence you have to pick in pair (char, int)and print it as per rule.

for(size_t i = 0; i < arr.size() - 1;) {

    int val = (int)arr[i+1] - 48;

    for(auto j = 0; i < val; j++) {

        cout << arr[i];

    }

    i += 2;

}

First for() is for traversing through whole arr.

Second for() is for printing the particular char for it's number of occurrences.

Note : Since the given arr is of type char you have to convert the second element of pair from char to int.

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

add a comment |

The way you have described the problem, the input will always in form of char followed by int. Hence you have to pick in pair (char, int)and print it as per rule.

for(size_t i = 0; i < arr.size() - 1;) {

    int val = (int)arr[i+1] - 48;

    for(auto j = 0; i < val; j++) {

        cout << arr[i];

    }

    i += 2;

}

First for() is for traversing through whole arr.

Second for() is for printing the particular char for it's number of occurrences.

Note : Since the given arr is of type char you have to convert the second element of pair from char to int.

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

add a comment |

The way you have described the problem, the input will always in form of char followed by int. Hence you have to pick in pair (char, int)and print it as per rule.

for(size_t i = 0; i < arr.size() - 1;) {

    int val = (int)arr[i+1] - 48;

    for(auto j = 0; i < val; j++) {

        cout << arr[i];

    }

    i += 2;

}

First for() is for traversing through whole arr.

Second for() is for printing the particular char for it's number of occurrences.

Note : Since the given arr is of type char you have to convert the second element of pair from char to int.

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

The way you have described the problem, the input will always in form of char followed by int. Hence you have to pick in pair (char, int)and print it as per rule.

for(size_t i = 0; i < arr.size() - 1;) {

    int val = (int)arr[i+1] - 48;

    for(auto j = 0; i < val; j++) {

        cout << arr[i];

    }

    i += 2;

}

First for() is for traversing through whole arr.

Second for() is for printing the particular char for it's number of occurrences.

Note : Since the given arr is of type char you have to convert the second element of pair from char to int.

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

edited Nov 22 '18 at 9:21

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

answered Nov 22 '18 at 9:10

Shravan40

4,30531631

add a comment |

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