invalid conversion from `char' to `const char*

bru · 05-08-2004, 10:14 PM

I'm writing a programm in C++ that mimics a database, and this problem keeps comming up, and I can't seem to get rid of it. I placed a comment above the line where the problem is. And sorry its so long!

Code:

#include <iostream>

#include <cstring>

#include <fstream>

#include <string>            

#include <stdio.h>

#include <iomanip>



using namespace std;



struct Score

{                   // making record fields

    char student[50];

    char studentscore[50];

    char student_name;

    int how_many;

};
.
.
.

// Add score function

void add_Score(int&, Score [])



{

	//declareing local varables that are needed



	char score[50];

	char studentscore[50] ;

	char student_name;

	int student_score ;

	int how_many;

	int new_score;

	int i = -1 ;

	int num_students;

	char choice = 'Y' ;

	Score The_Scores[50];



	cout << "You selected 'Add a score' to the list." << endl ;

	cout << endl ;



	// user enters a score which is stored in score[20])

	cout << "\t"<< "1) Enter the score (up to 3 characters) " 

		    << endl << "\t" << ": " ;



    fflush(stdin);

	gets(score) ;



//searches for the student name the user enters
//the following line is where the problem is 
	while (strcmp(score,  The_Scores[i].student_name)!=0 && i <=num_students)

		i++ ;



//if the student name was not found

	if (strcmp(score, The_Scores[i].student_name)!=0)

	{

		cout << "The student name was not found and will be added to the list." << endl ;

		

		cout << "\n\t" << "2) Please, enter the student's last name (up to 20 characters)" 

			<< endl << "\t" << ": " ;

		fflush(stdin) ; 

		gets(studentscore) ;

		

		cout << "\n\t" << "3) Now, enter the enter the students score: " 

			<< endl << "\t" << ": " ;

		fflush(stdin) ;

		gets(score) ;



	// the info is then stored in the array 'Scores' 

		strcpy(The_Scores[num_students].student, score);  

		strcpy(The_Scores[num_students].studentscore, score);

		strcpy(The_Scores[num_students].student_name, student_name) ;

		The_Scores[num_students].studentscore = studentscore ;

		The_Scores[num_students].how_many = how_many; 

		num_students++;  // increment after adding another score

	}



  // if there is a match with a student's name already in inventory, user is notified

  //if sutdents name is there option to update score is given

	{

		cout << setw(59) << "The student's name is already in the list." << endl ;

		cout << setw(70) << "Do you want to update the his/her Score (Yes/No)? " ;

		cin >> choice ;

		cout << endl ;





	if (choice == 'Y' || choice == 'y')

		{

			cout  << "Enter students new score: " ;

			fflush(stdin) ;		

			cin >> new_score ;

			The_Scores[i].student_name, score = The_Scores[i].student_name, new_score  ; //not sure if its right

		}

	}

	cout << endl ;

}	//end of function

Thanks for the help!

leonscape · 05-08-2004, 10:21 PM

studentname is a char, strcmp expects char*, so give it the address of student name.

Code:

while (strcmp(score,  &(The_Scores[i].student_name) )!=0 && i <=num_students)

And unless all your students of single character names (you've said upto 20 ), this ain't going to work.

bru · 05-08-2004, 10:23 PM

Thanks a million leonscape!!

One of these days I'll go from trying to learn C++ to kinda knowing C++

Thanks again!

leonscape · 05-08-2004, 10:30 PM

My advice, would be not to use char*, but use the string class from the STL. This would cure most of your problems. you've included the header why not use it?

if ( score != The_Scores[i].student_name && i <=num_students )

bru · 05-08-2004, 10:54 PM

Could you explain it a little more, I don't quite understand what you meen.

I'm still very new to programming, and trying to get an understanding of it all.

leonscape · 05-09-2004, 12:18 PM

char is a single character, ( as in studentname ) char* is a pointer to a char. arrays of chars are used for strings. a char array without an index is also a char*.

string is a c++ class. Its very useful, as unlike a char array, its dynamically sized ( no need to worry so much about its length. ) and can beused with operators.

eg.

Code:

string str2;
str2 = "hello world.";
if ( str2 == "hello" )
	str2 += "Adding letters";
str2 = str2 + "adding letters";

No need for all the extra functions. They work as you would expect. I just think it would be easier for your code.

bru · 05-09-2004, 03:07 PM

Ok I get it, thanks again leonscape!