invalid conversion from `const char*' to `char*'

deepinlife · 08-04-2006, 02:18 AM

guys
how can i elminate such error..

i use
g++ 3.3.6
kernel 2.4.31
distrubtion slackware 10.2

i always face this error and i solve it by make casting but i m asking about new soultion or some flags i pass to g++ to solve it?

vharishankar · 08-04-2006, 03:05 AM

Can you give the code you're trying to compile? Depending on what you're doing this might be an issue with the code rather than the compiler.

const char* is definitely not the same as char*

spooon · 08-04-2006, 03:09 AM

Yeah, it sounds like a violation of const-correctness.

deepinlife · 08-04-2006, 04:00 AM

Code:

char  meta_data::decode_int( const char* current ,int* charecter_lenght , long* integer )
{
	//make sure that it is integer
	if( *current != 'i' )
		return ERROR;
	//get the output of the integer
	const char *x = current+1 ;
	*integer = strtol( x , (char**)0 , 10 );
	;
	//check for underflow and overflow of strtol
	if( *integer ==  LONG_MAX || *integer==  LONG_MIN )
		return ERROR;
	
	//get  lenght
	while( true )
	{
		if(*(x++) == 'e')
			break;
	}
	*charecter_lenght = x-current ;
	return OK;
}

spooon · 08-04-2006, 04:22 AM

Seems fine to me. Can you show us the rest of the code and the exact error message and verify that the error is from this function? Maybe it's from the code that calls this function?

deepinlife · 08-04-2006, 06:21 AM

Code:

#include <iostream>

#include "../code/meta_data.h"
#include "../code/const.h"

using namespace std;

int main()
{
	meta_data  META;
	char array[] ="i55466466543e";
	int size=0;
	long lenght=0;
	char error = META.decode_int( array , &size ,&lenght);
	if( error!=ERROR)
		cout << "array =" << array<<endl
			<<"size= " << size <<"\t lenht = "<< lenght << endl;
	else cout<<"error\n";
	
	return 0;
}

Code:

class meta_data
{
	long lenght;

	public:
		char decode_int( const char* current ,int* charecter_lenght , long* integer );
};

deepinlife · 08-04-2006, 06:25 AM

i m sorry but i forgot to tell about the error message..
in the function defination
u see that i have declared
const char* x;
if i deleted the const word ,the error will appear..
invalid conversion from const char* to char*

vharishankar · 08-04-2006, 06:38 AM

Since you're using C++ I strongly recommend std::string instead of char* or const char*. You can always pass an std::string.c_str() to library functions which need const char* arguments and it's much less hassle to use std::string in all your programs since you needn't allocate and deallocate memory each time like you do for char*.

edit: since you're already using namespace std, it should be a matter of just declaring it as

string a_str = "A string";

deepinlife · 08-04-2006, 07:52 AM

thx Harishankar for the string advice..i will try to reconsider using char pointer , but i still want to know about the pointer problem now..

i know that c++ is so restrict about conversion but i need to know how to overcome it

Flesym · 08-04-2006, 09:35 AM

Ehmmm.., actually I don't see you problem:

Code:

const char *x = current+1 ;

You declared current as const char* so you have to declare x also as const, because there is no impicite conversion from const char* to char* (exactly what the error message told you). If you necessarily need to assign a "const" to a "non-const", then you can use the explicite convert operator const_cast:

Code:

const char *current;
char *x;
x = const_cast<char*>(current);

But be careful; very often you should consider to change your design instead of using const_cast. You may also want to go through some tutorials about type casting as well as the const modfier in C++. Simply ask google for these words.

PS. And as Harishankar said already: You should consider to use std::string

deepinlife · 08-05-2006, 12:53 AM

thx everyone .. i always use casting to solve this problem , but i m asking about another kind of solution..i mean is there a flag i pass to g++ which elminate reporting this kind of error?

spooon · 08-05-2006, 01:01 AM

Const correctness is a good thing. You are making a promise that you are not going to modify the stuff something points to, etc. So either you keep the promise, or you don't make it in the first place. What's so hard about that?

kalleanka · 08-05-2006, 04:40 AM

you declare: char array[] ="i55466466543e";

but call: meta_data::decode_int( const char* current ,int* charecter_lenght , long* integer )

and get: invalid conversion from `const char*' to `char*'

and for me it should be: invalid conversion from `char*' to `const char*'

It should not be a proble with const char to char but a problem with char to const char.

Strange?

kalleanka · 08-05-2006, 04:43 AM

maybe its her: const char *x = current+1 ;

maybe the "current" needs to be declared as a char.

spooon · 08-05-2006, 05:03 AM

Quote:

Originally Posted by kalleanka

you declare: char array[] ="i55466466543e";

but call: meta_data::decode_int( const char* current ,int* charecter_lenght , long* integer )

and get: invalid conversion from `const char*' to `char*'

and for me it should be: invalid conversion from `char*' to `const char*'

It should not be a proble with const char to char but a problem with char to const char.

Strange?

No, you can always do that. Do you understand what the const means? "const char *" means you cannot do anything that can modify the things pointed to by the pointer, which is more restrictive than "char *", so it's always okay to convert from "char *" to "const char *", but not the other way.