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

I'm having a problem with using character variables. I'm trying to make a char variable equal a string like "hello" but still be able to change it.

Code:

char arp;
int pra = 1;

if( pra == 1 ){
    arp = "hello";
}else
    arp = "bye";
}

Also i've tried this...

Code:

char arp[20];
int pra = 1;

if( pra == 1 ){
    arp = "hello";
}else
    arp = "bye";
}

your in c++ dude, use strings,

Code:

string arp;
int pra = 1;

if( pra == 1 )
{
    arp = "hello";
}
else
{
    arp = "bye";
}

dont forget to include <string>
also brackets are your friends

Oh lol...um good ideal. Thanks .

Explanation:

1)

Code:

char arp;
int pra = 1;

if( pra == 1 ){
    arp = "hello";
}else
    arp = "bye";
}

The line:

Code:

char arp;

declares a char variable. A char variable can hold a single character, e.g.

Code:

char arp;
arp = 'a';

...and a character in C++ is enclosed with single quotes--not double quotes. A string literal like:

"hello"

cannot be a char type.

2)

Code:

char arp[20];
int pra = 1;

if( pra == 1 ){
    arp = "hello";
}else
    arp = "bye";
}

First, your missing an opening bracket for your else branch. Then, in this line:

char arp[20];

you are declaring an array of characters. Generally, you cannot assign a string literal like "hello" to an array of characters. However, there is one little twist: with character arrays, you are able to assign a string literal to them at the time you declare them. For instance, you can do this:

char arp[20] = "hello";

When you do that the compiler will copy "hello" into the arp array character by character, and fill the remaining index positions with '\0' characters. You cannot subsequently do this, though:

arp = "bye";

because that doesn't fit the requirements: you are not assigning the string literal to arp at the time you declare arp. arp has already been declared.

Well, I don't know how hackish this is, but:

Code:

	char* arp = new char[20];
	int pra = 1;
	if( pra == 1 ){ arp = "hello"; }
	else  { arp = "bye"; }

That is if you still want to use this. In this way, only the memory addresses of "ehllo" or "bye" are assigned to the arp* pointer, and it all works. However, std::string is overpoweringly great, so whatev. I don't know the exact reason why you can't assign things like you did, I can't really think of how it would look on a low level...

Yes, very hackish.
Use the strcpy function in the <cstring> header.

Code:

strcpy(arp, "hello");

Code:

arp = "hello";

This is pointer assignment, and this is why you could not get it to work correctly. When you declare an array statically, it has a fixed position, and your named variable will always point to that location until it the array goes out of scope, and gets reclaimed by the system. What the above code tries to do is assign 'arp' to point to the string literal "hello" which is probably somewhere in read-only memory, but since you cannot reassign this pointer, you get an error.

Originally Posted by Tonto

Well, I don't know how hackish this is, but:

Code:

	char* arp = new char[20];
	int pra = 1;
	if( pra == 1 ){ arp = "hello"; }
	else  { arp = "bye"; }

That is if you still want to use this. In this way, only the memory addresses of "ehllo" or "bye" are assigned to the arp* pointer, and it all works. However, std::string is overpoweringly great, so whatev. I don't know the exact reason why you can't assign things like you did, I can't really think of how it would look on a low level...

No, no, no, no. That's bad. You've just created a memory leak by reassinging the address stored in the pointer from the memory you allocated initially using the new operator to a different address (the address of one of the string literals). Bad things could happen if you were to attempt to call delete on the pointer. If you don't belive me run this:

Code:

char * arp = new char[20];

cout << "Address stored in arp: " << hex << reinterpret_cast<int>(arp) << endl;
arp = "hello";
cout << "Address stored in arp: " << hex << reinterpret_cast<int>(arp) << endl;

On my machine I get this as output:

Code:

Address stored in arp: 320b50
Address stored in arp: 4170b0

As you can see, you lose the address of the memory you've initially allocated using the new operator.