problem with dynamic constructors, (cannot convert string literal to (char *) ).

Tarun Jha

#include
#include

using namespace std;

class String{
char *name;
int length;

public:
String(){ //default constructor
length = 0;
name = new char[length + 1];
}
String(char *s) //parameterised constructor
{
length = strlen(s);
name = new char[length + 1] ;

    strcpy(name, s);
}

void display(void){
    cout<

i should get output :
joseph
louis
lagrange
Joseph Louis
Joseph louis lagrange

but instead i am getting:
Joseph
louis
lagrange
louis
lagrange

Lost User

name = new char\[length + 1\];            //dyanamic allocation

strcpy(name, a.name);
strcpy(name, b.name);

You are copying both names to the same place instead of concatenating them using strcat. A better way would be something like:

name = new char\[length + 2\];            //dyanamic allocation
strcpy(name, a.name);     // copy the first name
strcat(name, " ");        // append a space character after the first name
strcat(name, b.name);     // append the second name

Jochen Arndt

Your join function is not joining but effectively setting only the second string:

void String :: join(String &a, String &b){
length = a.length + b.length ;
delete name;
name = new char[length + 1]; //dyanamic allocation

strcpy(name, a.name);
// This will overwrite name with b.name
//strcpy(name, b.name);
// You have to append b.name instead using strcat()
strcat(name, b.name);
//  or by copying to the current end
//strcpy(name + a.length, b.name);

};

cannot convert string literal to (char *)

To avoid that message change the constructor to accept a const char* argument:

String(const char *s)
{
length = strlen(s);
name = new char[length + 1] ;
strcpy(name, s);
}

CPallini

You know C++ standard library provides the string class, featuring the concatenation operator[^], don't you?

Tarun Jha

Thank you sir.. here's the working code

#include
#include

using namespace std;

class String{
char *name;
int length;

public:
String(){ //default constructor
length = 0;
name = new char[length + 1];
}
String(const char *s) //parameterised constructor
{
length = strlen(s);
name = new char[length + 1] ;

    strcpy(name, s);
}

void display(void){
    cout<

Tarun Jha

i am just beginning with c++, so did'nt knew. Thank you for informing

Tarun Jha

Thank you

Tarun Jha

Quote:

To avoid that message change the constructor to accept a const char* argument:

It worked as you told, but how and why it was giving error when i was simply using char, and how i will know in which conditions to use const char ? please can you elaborate

Jochen Arndt

Always use const for pointer and reference parameters when the content is not modified by a function. You have to use it when passing a string literal because that is by definition const (a "string literal" is a const char* which can't be assigned to a non-const char*).

Joe Woodbury

Tarun Jha wrote:

length = a.length + b.length ;

[Trivial] note that this actually should be:

length = a.length + b.length - 1;

(And please add a destructor. As a pure FYI, since I assume this is for learning hence not using std::string, you could use std::unique_ptr for name, avoiding the need for anything but a default destructor.)

Tarun Jha

thank you :)