std::string {aka std::basic_string<char>}' to 'char*' in assignment|

Question

tried to open a .cpp in code::blocks. Got few lines of error

Partial code :

void QSort(string List[], int Left, int Right)
{
  int i, j;
  char *x;
  string TEMP;

  i = Left;
  j = Right;
  x = List[(Left+Right)/2];

  do {
    while((strcmp(List[i],x) < 0) && (i < Right)) {
       i++;
    }
    while((strcmp(List[j],x) > 0) && (j > Left)) {
        j--;
    }
    if(i <= j) {
      strcpy(TEMP, List[i]);
      strcpy(List[i], List[j]);
      strcpy(List[j], TEMP);
      i++;
      j--;
   }
  } while(i <= j);

  if(Left < j) {
     QSort(List, Left, j);
  }
  if(i < Right) {
     QSort(List, i, Right);
  }
}

I recieve this error in line

 x = List[(Left+Right)/2];

cannot convert 'std::string {aka std::basic_string}' to 'char*' in assignment|

possible duplicate of How can I convert a std::basic_string type to an array of char type? — kfsone
– kfsone, Commented Nov 2, 2013 at 4:57
Why not use the lexicographical comparison operators for std::string? — Brett Hale
– Brett Hale, Commented Nov 2, 2013 at 5:04

kfsone · Accepted Answer · 2013-11-03 00:36:18Z

3

Because they're incompatible. You need to call a member of std::string which returns a const char*.

x = List[(Left+Right)/2].c_str();

Be aware: this pointer is only valid for the life time of the std::string or until you modify the string object.

This function returns a const char* so you'll need to change the definition of x from char* to `const char*.

const char* x;

or better still, remove that line, and combine the two

void QSort(string List[], int Left, int Right)
{
    string TEMP;

    int i = Left;
    int j = Right;
    const char* x = List[(Left+Right)/2];

Infact, here's a rewrite that uses standard C++ algorithms throughout (std::string::compare instead of strcmp). This may make it easier for you to focus on the algorithm itself.

void QSort(string List[], int Left, int Right)
{
    int i = Left;
    int j = Right;
    const int mid = (Left+Right) / 2;

    for (;;) // repeat until we break.
    {
        // write both comparisons in terms of operator <
        while (List[i].compare(List[mid]) < 0 && i < Right)
            ++i;
        while (List[mid].compare(List[j]) < 0 && Left < j)
            --j;
        // if i == j then we reached an impasse.
        if (i >= j)
            break;
        std::swap(List[i], List[j]);
    }

  if(Left < j)
    QSort(List, Left, j);

  if(i < Right)
    QSort(List, i, Right);
}

edited Nov 3, 2013 at 0:36

answered Nov 2, 2013 at 4:56

kfsone

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5 Comments

Bernardo Baalen Over a year ago

I think is compiler issue, many lines of code affected.. any other solution?

kfsone Over a year ago

It's not a compiler issue. You can't convert a std::string to a const char* because they are incompatible. You need to call the c_str() method of string to get the const char* which the std::string is managing.

kfsone Over a year ago

Don't let the name, string, confuse you: std::string is a class name. char foo[] = "hello"; and char* foo = "hello"; are C-style strings, while std::string foo = "hello"; is a different beast entirely.

kfsone Over a year ago

std::string::c_str() returns a const char* - the const indicating "you can't modify the thing the pointer points to". You just need to change x from char* x; to const char* x;. Infact, remove the char *x; line and make the change shown above.

kfsone Over a year ago

Take a look at my cleanup of the algorithm which avoids using char*-related functions entirely and uses the std::string member functions to do the compares.

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