C++ arrays and strstrok





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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;
}









share|improve this question























  • 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




















2















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;
}









share|improve this question























  • 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
















2












2








2








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;
}









share|improve this question














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;
}






c++ arrays strtok typed-arrays






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share|improve this question










asked Nov 22 '18 at 7:52









Be BehindBe Behind

197




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





















  • 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





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







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














2 Answers
2






active

oldest

votes


















2














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;
}





share|improve this answer































    2














    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;
    }



    1. First for() is for traversing through whole arr.

    2. 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.






    share|improve this answer


























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      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;
      }





      share|improve this answer




























        2














        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;
        }





        share|improve this answer


























          2












          2








          2







          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;
          }





          share|improve this answer













          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;
          }






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 22 '18 at 9:01









          Aurelien NormandAurelien Normand

          491




          491

























              2














              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;
              }



              1. First for() is for traversing through whole arr.

              2. 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.






              share|improve this answer






























                2














                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;
                }



                1. First for() is for traversing through whole arr.

                2. 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.






                share|improve this answer




























                  2












                  2








                  2







                  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;
                  }



                  1. First for() is for traversing through whole arr.

                  2. 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.






                  share|improve this answer















                  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;
                  }



                  1. First for() is for traversing through whole arr.

                  2. 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.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 22 '18 at 9:21

























                  answered Nov 22 '18 at 9:10









                  Shravan40Shravan40

                  4,44531633




                  4,44531633






























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