Char transformation on simple string array on C-CodePudding

Please help to understand what's wrong with my simple C code

  #include <iostream>
  #include <cstring>
  #include <string>

  using namespace std;

  int main()
  {
      string s = "qwertyuiopasdfghjklzxcvbnm";
      string match = "w";
   
      int n = s.length();
   
      char char_array[n   1];
   
      strcpy(char_array, s.c_str());

   
      for (int i = 0; i < n; i  )
          if (match.compare(char_array[i]) == 0) {
              cout << char_array[i]; 
          }
      return 0;
  }

I receive an error :

error: invalid conversion from ‘char’ to ‘const char*’ [-fpermissive]

Please help with this conversion char to *char and compare them correctly

CodePudding user response：

There are 2 mistakes in your given example.

Mistake 1

You have written the statement:

 char char_array[n   1]; //since n is not a compile time constant so this is not standard C

In C , the size of an array must be a compile time constant. So you cannot write code like:

int n = 10;
int arr[n];    //incorrect

Correct way to write this would be:

const int n = 10;
int arr[n];    //correct

Mistake 2

You are trying to convert char to const char* as the error says in the compare() method.

If you just want to find out whether a given character occurs in a std::string then there are 2 options/solutions(maybe more).

Solution 1

 #include <iostream>
  #include <string>

  using namespace std;

  int main()
  {
      string s = "qwertyuiopasdfghjklzxcvbnm";
      char match = 'w';  //use char instead of std::string
   
      for (int i = 0; i < s.length(); i  )
          if (s[i] == match) {
              cout << "element: "<< match <<" found at index: "<<i<<std::endl; 
          }
      return 0;
  }

As shown in the above solution 1 you don't need to create a separate array.

Solution 2

You can use std::string::find to find a given substring inside another string. This would look like:

 #include <iostream>
  #include <string>

  using namespace std;

  int main()
  {
      string s = "qertyuiowpasdfghjklzxcvbnm";
      std::string match = "w"; //using std::string instead of char
   
      std::size_t index = s.find(match);
      
      if(index != std::string::npos)
      {
          std::cout<<"string: "<<match<<" found at index: "<<index<<std::endl;
      }
      else 
      {
          std::cout<<"string: "<<match<<" not found"<<std::endl;
      }
      return 0;
  }

The output of solution 2 can be seen here.

Solution 3


#include <iostream>
#include <string>
int main()
{
   std::string s = "qertyuiopasdfwghjklzxcvbnm";
   std::size_t index1 = s.find_first_of('w');

   if(index1 != std::string::npos)
   {
        std::cout<<"found at index: "<<index1<<std::endl;
  }
    else 
    {
        std::cout<<"not found"<<std::endl;
        
    }
    return 0;
}

CodePudding user response：

As it states in the error, you cannot convert a single character to a string. But you can use the std::string(size_t , char ) constructor.

#include <iostream>
#include <cstring>
#include <string>

using namespace std;

int main()
{
    string s = "qwertyuiopasdfghjklzxcvbnm";
    string match = "w";
 
    int n = s.length();
 
    char char_array[n   1];
 
    strcpy(char_array, s.c_str());

 
    for (int i = 0; i < n; i  )
        if (match.compare(string(1, char_array[i])) == 0) {
            cout << char_array[i]; 
        }
    return 0;
}