Writing an enum in a struct to a struct array

[timmio888]

So I have a project due tomorrow (yes, this is homework, but only a clarification/technical question that I can't find on Google) and, as the title says, I'm trying to write from a file to a struct array and store them as structs.

In addition, I don't know how to access the "height" and "width" ints in the struct.
Here is my code. The line denoted with asterix's is the part I'm inquiring about.

#include <iostream>
#include <string>
#include <fstream>
using namespace std;

struct art {
	string artist;
	string title;
	enum medium {a, o, p, f, w };
	struct size {
		int height;
		int width;
	};
	enum room {m, g, u, n, s, e, b };
	int price;
} ArtWork, GalleryCollection[500];

int main()
{
	ifstream input;
	input.open("art.dat");
	while (input >> GalleryCollection[])
            {
	int j=0;
        input >> GalleryCollection[j].artist >> GalleryCollection[j].title >>

****************************************************************************
                 GalleryCollection[j].medium >>

                 GalleryCollection[j].ArtHeight.height >>

                 GalleryCollection[j].ArtWidth.width >>

                 GalleryCollection[j].room >>
****************************************************************************
                 GalleryCollection[j].price >> endl;
		j++;
	    }
	input.close();

	system("pause");
	return 0;
}

Since "medium" and "room" are of type enum, it won't let me pass it to a struct array. I have no idea how to access the enum type(s) within the "art" struct, nor how to access the objects inside the "size" struct within the "art" struct. Any thoughts?

[AbstractionAnon]

medium and room are enum declarations. They do not allocate a member variable in the struct type art.

Move the enum declarations out of the struct declaration and add an enum member variable in their place.

enum medium_t {a, o, p, f, w };
enum room_t {m, g, u, n, s, e, b };
struct art 
{  string artist;
   string title;
   medium_t medium;
   struct size 
   {  int height;
       int width;
   };
   room_t  room;
   int price;
} ArtWork, GalleryCollection[500];

PLEASE USE CODE TAGS (the <> formatting button) when posting code.
It makes it easier to read your code and also easier to respond to your post.
http://www.cplusplus.com/articles/jEywvCM9/
Hint: You can edit your post, highlight your code and press the <> formatting button.

[timmio888]

Oh okay, that makes much more sense.

One more thing though: how do I call the declarations "height" and "width" inside struct "size" within struct "art?" Would it be something like Artwork.size.height?

[AbstractionAnon]

You don't "call" declarations.

Line 7 (my previous post): You have the same problem here with your struct as you did with your enums. size is a struct declaration. It is not a member variable. Sorry about not catching that in my previous post. Same solution applies.

enum medium_t {a, o, p, f, w };
enum room_t {m, g, u, n, s, e, b };
struct size_t
{  int height;
    int width;
};
struct art 
{  string artist;
   string title;
   medium_t medium;
   size_t  size;
   room_t  room;
   int price;
} ArtWork, GalleryCollection[500];

Would it be something like Artwork.size.height?

Yes.

[TarikNeaj]

Learn to use enum class instead of plan enum -

http://www.cprogramming.com/c++11/c++11-nullptr-strongly-typed-enum-class.html

http://stackoverflow.com/questions/18335861/why-is-enum-class-preferred-over-plain-enum

[timmio888]

Thanks so much for the feedback!

Unfortunately, when you fix one problem, two more pop up. I'll just paste the whole program because this new problem is a little hard to explain out of context.

#include <iostream>
#include <string>
#include <fstream>
using namespace std;

enum mediumt {a, o, p, f, w};
enum roomt {m, g, u, n, s, e, b};
struct sizet {
	int height;
	int width;
};
struct art {
	string artist;
	string title;
	sizet size;
	mediumt medium;
	roomt room;
	int price;
} ArtWork, GalleryCollection[500];

int main()
{
	ifstream input;
	input.open("art.dat");
	

	{
		int j=0;
		input >>
			GalleryCollection[j].artist >>
			GalleryCollection[j].title >>
*******************************************************
			GalleryCollection[j].medium >>
*******************************************************
			GalleryCollection[j].size.height >>
			GalleryCollection[j].size.width >>
*******************************************************
			GalleryCollection[j].room >>
*******************************************************
			GalleryCollection[j].price;

		cout << "The name of the artist is " << GalleryCollection[0].artist << endl;
		cout << "The name of the painting is " << GalleryCollection[0].title << endl;
		cout << "The material the painting was made out of is " << GalleryCollection[0].medium << endl;
		cout << "The height of the painting is " << GalleryCollection[0].size.height << endl;
		cout << "The width of the painting is " << GalleryCollection[0].size.width << endl;
		cout << "The room where the painting is hung is the " << GalleryCollection[0].room << " room" << endl;
		cout << "This painting is worth $" << GalleryCollection[0].price << endl;
	}
	input.close();

	system("pause");
	return 0;
}

A bit of background: I have a txt file called art.dat which contains "TedSmith RedRock o 10 12 n 12500." That's <artist name> <painting title> <material painted with> <height> <width> <room stored in> <price>.

For a reason unknown to me, the asterix'd lines of code "GalleryCollection[j].medium" and GalleryCollection[j].room" are giving me errors, saying that "no operator matches these operands ">>" on line 32 (GalleryCollection[j].title >>). However, if I take those two lines out, the program runs fine. How do I fix it?

[Thomas1965]

mediumt and roomt are not build in types and therefore the operators >> and << do not work.

One way to fix it is to read the input into an int and convert it.

int medium;
input >> medium;
GalleryCollection[j].medium = (mediumt) medium;

Alternatively you could write your own insertion operator, but this is more complicated.

[AbstractionAnon]

"TedSmith RedRock o 10 12 n 12500."

n and o are char data.

Thomas1965's suggestion to read into an int is not going to give you the results you expect. Attempting to read a 'o' or 'n' into an int is going to cause the >> operation to fail.

[Thomas1965]

AbstractionAnon is right. Somehow I failed to notice that they are chars. So it might better to read them in a char and convert it to mediumt and roomt types.