invalid conversion from `char*' to `char'

Hello, I am new to C++, I just started yesterday. I am trying to make a fuction where I can return a line from a .txt with linesi(file.txt,line number).

But I get the following errors, and being new, I have no idea why:

C:\Documents and Settings\Family\Desktop\Untitled21.cpp In function `int main()':
17 C:\Documents and Settings\Family\Desktop\Untitled21.cpp invalid conversion from `char*' to `char'
17 C:\Documents and Settings\Family\Desktop\Untitled21.cpp initializing argument 1 of `int linesi(char, int)'
C:\Documents and Settings\Family\Desktop\Untitled21.cpp In function `int linesi(char, int)':
22 C:\Documents and Settings\Family\Desktop\Untitled21.cpp invalid conversion from `char' to `const char*'
22 C:\Documents and Settings\Family\Desktop\Untitled21.cpp initializing argument 1 of `std::basic_ifstream<_CharT, _Traits>::basic_ifstream(const char*, std::_Ios_Openmode) [with _CharT = char, _Traits = std::char_traits<char>]'
28 C:\Documents and Settings\Family\Desktop\Untitled21.cpp invalid conversion from `char*' to `int' 24 C:\Documents and Settings\Family\Desktop\Untitled21.cpp [Warning] address of local variable `buffer2' returned


Just wish I knew what I was doing wrong. Here is my code:

Code:

#include <iostream>
#include <fstream>
using namespace std;

int linesi (char b, int c);
 
int main ()
{
   char b[256];
   int c;
   char buffer[256];
   cout << "Please enter the filename of a .txt file: ";
   cin >> b;
   cout << "Please enter the line number: ";
   cin >> c;
   ofstream lfo("list2.txt");
   lfo << linesi(b,c);
}
     
int linesi (char b, int c)
 {
     ifstream ag(b);
     int x = 1;
     char buffer2[256];
   while (!ag.eof())
   {
    ag.getline(buffer2, 200);
    if (x == c) { return buffer2; }
    x++;
}
}

I've been working on this for hours.. help would be greatly appreciated.

Maybe something along this line...

Code:

#include <iostream>
#include <fstream>
using namespace std;

char *linesi (char *b, int c)
{
   ifstream ag(b);
   for ( int x = 1; ag.getline(b, 200); ++x )
   {
      if ( x == c )
      {
         return b;
      }
   }
   return 0;
}

int main ()
{
   char b[256];
   int c;
   cout << "Please enter the filename of a .txt file: ";
   cin >> b;
   cout << "Please enter the line number: ";
   cin >> c;
   ofstream lfo("list2.txt");
   lfo << linesi(b,c);
}

/* my output
Please enter the filename of a .txt file: testpp.cpp
Please enter the line number: 5

H:\test>type list2.txt
char *linesi (char *b, int c)
*/

Much appreciated!

Hi,

In C++, the type of the variable you send to a function must exactly match the type of the variable the function is expecting. You declared your function like this:

Code:

int linesi (char b, int c);

The function is expecting a char type and an int type. You called your function like this:

Code:

char b[256];
int c;
..
...
linesi(b,c);

You are sending your function a char array and an int. A char array is not the same thing as a char type. Here is how you declare both a char array and a char type:

char b[256];
char ch = 'x';

If you sent the variable ch to your function instead of b, you wouldn't get an error. It turns out that for a char array, the brackets and the size are part of the type of the variable. Your variable 'b' is actually of type char[256]. Here is an example:

Code:

#include <iostream>
#include <fstream>
using namespace std;

int linesi (char b[256], int c, char ch);
 
int main ()
{
   char b[256];
   int c;
   char ch='x';
   
   cout << "Please enter the filename of a .txt file: ";
   cin >> b;
   cout << "Please enter the line number: ";
   cin >> c;
   
   linesi(b, c, ch);

   return 0;
}
     
int linesi (char b[256], int c, char ch)
{
	
	cout<<endl<<"Here is your input:"<<endl;
	cout<<b<<endl;
	cout<<c<<endl;
	cout<<ch<<endl;
	
	return 100;
}