Ashley - being very familiar with this bug, any image will do. Or not. An image will upload successfully one moment, then not the other. The likelihood of the image being an issue is, having observed this error for a long time, 0. I could upload some images to make you happy, but all of them would as likely upload succesfully on my system as they would crash C2, so I don't much see the point.

You also don't get access to any settings options. It hangs before getting to the dialog where you input the spritesheet cell dimensions. I have observed occasions where the hang has solved itself after a few minutes and I am shown the dialog where cell dimensions may be entered - although these have been rare. More often than not I give it two minutes, then force-close and re-open C2.

I can't *intentionally* reproduce this bug as it seems to happen entirely at random, but has often happened in the past, and is happening on my current project all the time as I'm making heavy use of the sprite-sheet importer. Ashley : I've found this bug reported several times now, and always dismissed because it could not be reproduced - when it has in fact been reproduced by many users. Could it please be investigated?

I'm jamming at present and manually slicing up all my sprite-sheets is really cramping my time-frames - but it's necessary as it takes less time than the 'force close ten times in a row restart and cross my fingers for attempt 11' importing method. I know a fix won't come by the jam end or anything but it'd be nice to have this problem looked at before the next one!

brent_hamel

This shader is for sprite; to rotate transform skew and much more .. still wip ..

http://gigatron3k.free.fr/html5/C2/FX/sutility/

Love this plugin, and really want to get it working on sprites - is there anywhere to download it? I hope so! <img src="{SMILIES_PATH}/icon_e_biggrin.gif" alt=":D" title="Very Happy">

Congratulations! To check diagonally you'll need to add to that 'OR' condition (I didn't realise diagonals counted in gomoku!). You see how it currently checks at X/Y +/-1, like so:

Array.At( Array.CurX - 1 , Array.CurY )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX + 1 , Array.CurY ) = "yellowyellowyellow"[/code:162de7n3]

Note these lines check orthogonally. For example, X-1, X, X+1 describes a horizontal line. You'll have to add new checks to the OR expression which describe diagonal lines. Here's a starter, a diagonal line running top-left to bottom-right:

[code:162de7n3]Array.At( Array.CurX - 1 , Array.CurY - 1 )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX + 1 , Array.CurY + 1 ) = "yellowyellowyellow"[/code:162de7n3]

Is that a 9-Patch?

Add a condition to the action in which troops are purchased which checks to see if the player has enough points. If the points are stored as a variable, they can be compared using System > Compare variable. If they are stored on an object's variable, you will have to use Object > Compare variable.

Make sure it checks if their points are greater than OR equal to the cost!

Good luck - if you have any problems make sure to post your capx - it makes it a lot easier to see what you mean and how to fix it

I think the simplest way to do this might be to have the camera track an object the size of the screen. Have this object always pathing towards the player. When it overlaps areas that shouldn't be seen by the camera, steer it away from those objects. Good luck!

To do this you'll want loops, an an array will make it easy too. Add an array to your project with a Width of 5, a Height of 5, and a Depth of 1. The array can be seen as a 'spreadsheet' where you store the result of each turn. When a player places a piece, fill the appropriate cell in the array with the appropriate value. For example:

The YELLOW player's turn, they have placed a piece at X:
.....
.YYX.
..B..
..B..
.....

Array > Set at x,y > X: 3, Y: 1, Value: "yellow" [/code:2s1fo3nl]

Running this at the end of each turn will make it much easier to check for a win-state - which is what we check for next. There are a lot of ways to do this, but here's what I can think of:

[code:2s1fo3nl]
Array > For each element (xy)
System > Compare two values > 
First Value: Array.At( Array.CurX - 1 , Array.CurY )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX + 1 , Array.CurY ) = "yellowyellowyellow" | Array.At( Array.CurX , Array.CurY + 1 )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX , Array.CurY - 1 ) = "yellowyellowyellow" | Array.At( Array.CurX - 1 , Array.CurY )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX + 1 , Array.CurY ) = "blueblueblue" | Array.At( Array.CurX , Array.CurY + 1 )&Array.At( Array.CurX  , Array.CurY )&Array.At( Array.CurX , Array.CurY - 1 ) = "blueblueblue"
Second Value: 1
[/code:2s1fo3nl]

This seems pretty long-winded but it's actually a single event, which is nice. The 'For each element (xy)' action means the event will be run for every cell in the Array's x,y dimension. The first of the two values to compare is a long 'or' expression - each vertical line '|' character can be read as 'or'. The conditions in the expression create strings that represent the orthogonal lines through the current cell, and compare them to strings that define a win-state, eg: "blueblueblue" or "yellowyellowyellow". Because they are part of an 'or' expression, if any of the conditions in this expression are met, the output of the expression will be "1" - so the second value with which to compare the first should just be "1".

Place whatever actions go in your win condition into this action - good luck! What a fun little problem.

The 'faster than its supposed to' thing is because of the density - which is judged by the area of the collision polygon. By reducing the area of the polygon from the bounding box, you decrease the perceived 'mass' of the object.

Unfortunately, I can't explain why you're getting asymmetrical movement problems - it's not something I've come across myself

In the most basic terms:

Ball.Y > BarLEFT.Y

? Add ( dt * x ) to BarLEFT.Y

: Subtract ( dt * x ) from BarLEFT.Y

Post a CapX, this will make it much easier for people to diagnose and solve your problem.

And since your English is bad I'd recommend a 'thank you' is customary when someone volunteers help