Please help me with a strcat() error

[Koli]

here is my main:

int main(int argc, char* argv[])
{
char cA, cB;
cA = 'A';
cB = '\x6';
cA = strcat(cA, cB);
cout << cA;
return 0;
}

I'm trying to add the cA and the cB character together...does anyone know what I'm going wrong, when I get the following error message?:

cannot convert parameter 1 from 'char' to 'char *'

I'm using Visual C++ 6.

Thanks!

[Xterria]

You can't do this because the type 'char' represents only one character. Which means if you put cA with the letter 'A' its ok but if you put it to 'AB' its not. To get around this you have to make your chars arrays. this is what you should have:

int main(int argc, char*argv[])
{
      char cA[30];     //Notice that now you can put 30 chars in
      char cB[30];     //Ya can with this, too
cA = 'A';
cB = '\\x6';     //put the extra \ so that the compiler prints one
strcat(cA, cB);  //You don't need cA =
cout << cA;
return (0);
}
/* This should print: 'A\x6' */

I hope this helps you. If it doesn't just tell me

[lightatdawn]

>cannot convert parameter 1 from 'char' to 'char *'

That would be your problem, yes. The error message tells you exactly whats wrong. cA needs to be of type "char *" (a pointer to a character string) not of type char. You need to declare your variables like this:

char *cA, *cB;

/* Edit */
I see someone replied before i finished typing. One note: You only want that extra "\" if you want to display the next "\" instead of the escape sequence that i believe you intended here.

Also: the cA[30] method is also acceptable for this but why declare room for 30 characters when you're only using 3 or so? Use a pointer and "new" to declare only as much space as you'll need.

[Koli]

Actually, all I want to do is add the ASCII value 6 which is the Spade to a character, any Ideas?