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I have a doubt about what data structure I have to use to store the map

Why not use the same data structure as the nav_msgs/OccupancyGrid ?

This structure has a field `data`

, which is a 2D-Matrix and another field, nav_msgs/MapMetadata which is used to store the origin, width, height and resolution of your map.

But I don't know where exactly is the robot on the world that I have to map

Since you are doing the mapping of the envirronment you are free to choose any origin for your robot, as long as you keep track of all your obstacles relatively from your origin.

Using a vector, or a 2D matrix, here is a problem, because, vector and matrices indices start at 0, and there could be more room behind the robot to map.

True, the first index is always `0`

, but it's your choice to assign the coordinates to the indexes, meanning that index `0`

isn't necessarily the coordinate `(0;0)`

.

All that being said, you can decide when you start your mapping to set the current position of the robot to be the coordinate `(0;0)`

, for the example I choose to set the map origin to the bottom right corner of the map. I would calculate the origin coordinates according to your laser range. I would set the `origin`

, `width`

and `height`

to ensure that everything the robot can "see" at its start position could be stored in your map, so I would choose something like

```
origin.x = -(laser_range + 1)
origin.y = -(laser_range + 1)
height = 2*(laser_range + 1)
width = 2*(laser_range + 1)
```

The origin are negatives to be at the bottom right and all the values are incremented by 1 to be greater than the laser range.
From that, all your cells are the indexes of your matrix. From that and knowing your resolution is 1 meter, if your laser has a range of `4 meters`

you get a matrix of size `100`

. Your first index being the origin of your map.

Now every index can be calculate from any coordinates :

```
index = floor((x - origin.x)/resolution) + floor((y - origin.y)/resolution)*width
```

The last `*width`

is because you have a 2D-Matrix, if you browse your map from bottom to top and right to left, one increment of `y`

is equal to an increment of `width`

in your vector.

With your exemple if you see an obstacle two meters in front of the robot, you know that your are two meters in front of `(0;0)`

meaning `(2;0)`

. That represent the index :

```
index = floor((2-(-5))/1) + floor((0-(-5))/1)*10
index = floor(7) + floor(5)*10
index = 57
```

If your robot has moved don't forget to add the translation in the calculation :

```
index = floor((x + robot.x - origin.x)/resolution) + floor((y + robot.y - origin.y)/resolution)*width
```

Now if you detect obstacles further than your map limits you just have to recalculate a new origin and increase the size of your map. All your previous indexes can be recalucated using conversion from the actual coordinates of the robot and then another conversion to with the new data of your map.

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