How to "push" points to a list

Hello. I would like to know how I set in Points(x,y) to a list. Like this:

Code:

---

#include <list>

//Defining point type
typedef struct Point{
        int x, y;
}Point;

using namespace std;
list<int> lista;
lista.push_back(???);

---

Any ideas how it should look like?

Maybe

Code:

---

list<Point> lista;
Point foo;
lista.push_back(foo);

---

I´m sorry. I started with C++ very recently and dont know how to handle the typedef. Have tried all I can come up with but it still dosent work. What would you set "foo" to for example?
Just some things I have tried:

Code:

---

Point 1.x;
Point 1 2;
Point 1;
Point (1,2);
Point (1.x,2.y);

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and so on. Any clue? =)

You could probably do

Code:

---

lista.push_back(Point(1, 2));

---

Or you can do:

Code:

---

Point foo(1,2);
lista.push_back(foo);

---

--
Mats

Code:

---

Point foo(1,2);

---

gives:

Code:

---

error C2078: too many initializers
error C2440: 'initializing' : cannot convert from 'int' to 'Point'

---

and

Code:

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lista.push_back(Point(1, 2));

---

gives:

Code:

---

error C2661: 'Point::Point' : no overloaded function takes 2 arguments

---

My typedef is a bit odd for a point. Guess the best would be to change it maybe? Tried:

Code:

---

typedef struct Point{
        (int x, int y);
}Point;

---

But your suggestions still gives errors =(

If the struct is just a struct, the initialization is

Code:

---

Point foo = { 1, 2 };
lista.push_back(foo);

---

Ah, ues, of course. You need a constructor for Point to do that.

Either do:

Code:

---

Point foo = { 1, 2 };

---

or

Code:

---

struct Point{
        int x, y;
        Point(ax = 0, ay = 0): x(ax), y(ax) {};
};

Point foo(1, 2);

---

Note that in C++ the typedef struct Point {} Point is unneccesary, you can use Point straight from the struct declaration, the typedef is implicit.

--
Mats

Tons of thanks. Seems to work great =D

Just one more small question: is it possible to get the x and y values from this?

Code:

---

Point foo = { 1, 2 };
list.push_back(foo);
// tried this but dosent seems to be the correct way
int x = (list.pop_back()).x;

---

edit: solved with
int x = lista.back().x;