Advice B) If something isn't working, trying a simpler version of it, and build upwards. In this case, if "Blit every tile of every layer" isn't working, try a "Blit every tile of the first layer only" for testing.
std::iterator uses [] to support dereferencing. It looks like you are forgetting to de-reference it, and are trying to use it to select the layer or the tile.
std::vector<int> myVector; std::vector<int>::iterator it = myVector.begin(); it[3] //Accesses the fourth element after whereever "it" is pointing to. it++; it++; it[3] //Accesses the 6th element.
myVector.begin() = 0
++ = 1
++ = 2
[3] = 5 (2 + [3] = 5)
It should be:
std::vector<myTile>::iterator tmpIt; tmpIt = (*it)[X+(Y * MAP_WIDTH)].begin(); //De-reference 'it' first!
But really, for compact loops like that, it's better to break things out clearly, like this:
std::vector<myTile> tiles = (*it)[X+(Y * MAP_WIDTH)]; std::vector<myTile>::iterator tmpIt; tmpIt = tiles.begin();This makes it easier to see and debug where the mistakes are.
However, "tmpIt" is a meaningless name. All your iterators are temp, and all of them are iterators, so "temporary iterator" doesn't describe it at all. You might as well call all your integers, "holdsANumberOfSomeKind", because neither is descriptive.
Better (more descriptive) names make it less likely for mistakes to occur, and easier to spot them when they do occur.
Furthermore, your loop is three loops deep, plus an embedded if() statement. This calls greatly for you to break out the function into more than one function for clarity, for ease-of-debugging, and for ease-of-reading.
void DrawArea()
{
for(...each layer in area...)
{
DrawLayer(layer);
}
}
void DrawLayer()
{
for(...every tile in layer...)
{
DrawTile(tile);
}
}[Edit:] Ninja-d by two people saying the exact thing I'm saying, and then the OP responding. Posted anyway (despite seeing the "2 people posted while you were commenting" popup), for additional comments and explanations - and because I took the time to type it all out, and am not letting it go to waste.
