Basic C++ function, can I make it work?

Question

So, I'm reading Schildt's book 3rd edition about C++ and I'm doing all examples, but I have some PHP background and when I tried some stuff it occurs that it can not be compiled this way.I saw the Schildt's solution, so I'll give what I've tried to do and how it's done in the book, what I need to know, is there any way to make it work adjusting my function?

Here's what I'm trying

class card {
char author[40];
//char book[30];
int count;
public:
    void store(char *auth,int ct);
    void show();
};
void card::store(char *auth,int ct){
    &author = *auth;

    count = ct;
}
int main(){
    card ob1, ob2;
    ob1.store('Tolkin',10);
    ob2.store('Pratchet',3);
    ob1.show();
    ob2.show();
return 0;
}

And here's the Schildt's solution:

class card {
char author[40];

int count;
public:
    void store(char *auth,int ct);
    void show();
};
void card::store(char *auth,int ct){
    strcpy(author, auth);

    count = ct;
}

Answer 1

The quick fixes:

• Instead of char author[40] use std::string.
• store(const std::string& auth,int ct)
• author = auth; (std::string has assignment operators)
• ob1.store("Tolkin",10); (single quotes are for char-literals)
• Give card::show() a body. You currently have just the declaration. And because show() does not mutate card, make it a const member function: void show() const;

The real fix (sounds lapidar, but is my serious, well-intentioned advice):

• Get a good introduction to C++.

Answer 2

Not really, no. For any given instance of card, author is stored at a specific, fixed, memory location relative to the rest of the object. (author is actually inside the object: if you do std::cout << sizeof(card) << std::endl; you'll see that card is 40 bytes, because all it contains is an array of forty characters.) &author = ... is trying to tell the compiler that author is now stored somewhere else, but there's no way to tell the compiler that, because you already promised that author is stored inside the card.

That said, you could change your declaration of author to be a true pointer, rather than an array:

char * author;

and then assign to it like so :

author = auth;

but that's not a good idea. Whenever you pass a pointer around, you have to keep track of where it's gone, to make sure that all of your pointers are always valid, and that you never lose a pointer to memory you need to de-allocate later.

Answer 3

I think the basic misunderstanding here is the way that C/C++ handle "char arrays" as strings. The strcpy routine copies the contents of a string, where the = assignment operator (applied to char) copies a single character, or a pointer to the string. &author = *auth will look at the auth pointer, dereference it using *, and take the single char found there, then take the address of (&) your char[] named author, and try to change the address to the char value.

You could…

• Use strcpy to copy the contents of the string (but, in new code, don't use strcpy, use strncpy instead!)
• Store a pointer to the string provided (char* author in your class would be assigned as author = auth; but then, if auth is free()d or delete[]()ed later, you will have a pointer to memory that no longer contains your string, which is bad)
• Use a C++ std::string object instead of a C-style char[] for your string, which will behave more like a PHP string would. std::string author could be copied from std::string auth using author = auth.

String-handling in C++ is a big subject, but you will definitely want to get a good understanding of the differences between "thing" and "pointer-to-thing" types … !

Also, in C++, you must use "" around strings, and '' around single chars. There is a lot less "magic" in a C/C++ "" string, though; only \x type escape sequences work (for example, there is no "$var" substitution available). In Perl/PHP/Bourne/... you use '' for non-escaped strings and "" for escaped strings; in C++, since char and char[]/std::string are different types, they use the punctuation differently.

Answer 4

you are mixing up some types: PHP is not typed while c++ is.

you can t compile because you are trying to assign a pointer to a reference.

&author = *auth;

i suggest to read a lot of documentation about reference and pointers!! cheers!