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I have in mind coding a multiplayer game with a similar combat to zelda's. Since i will use flash, i'm limited to use TCP. My question is, is this goal possible? It's not as fast-paced as some shooters, but i'm not sure if it's too much for tcp?

tl;dr: TCP for zelda-like multiplayer combat style?

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  • \$\begingroup\$ Which Zelda game in the series? The combat is very different between Top-down (e.g. Zelda 1), Side-view (Zelda 2) and the 3D Zeldas. \$\endgroup\$
    – msell
    Commented Dec 13, 2012 at 6:28

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Why don't you try it out in a simple test project? TCP is a data stream, to avoid clogging up the stream (the "packets" will sit there untill they are read) you should call the receive function untill there is no data left to read. If you can sent 1024 bytes packages every frame without to much delay, then you should be good.

note: I know packet/package is not the right terminology for TCP, but I couldn't think of the right way to describe it.

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  • \$\begingroup\$ That's what i did, but i'd appreciate some feedback in the subjet. A real test would take making the actual game, so i'm looking for some feedback before being on it. \$\endgroup\$ Commented Dec 13, 2012 at 11:17
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    \$\begingroup\$ A real test, on the subject of networking would be to estimate the number of times you are going to send data and the size of the data you are going to send, then make a simple application that will go a bit over those values. If everything runs fine then you know it's the way to go, if you have some issues you can try to solve the problem and test again, or search for another method. You want to keep track of ping etc as well btw, and you'll need to test over the internet or at least a very busy network (turn on some downloads) to actually stresstest it properly. \$\endgroup\$
    – Kevin
    Commented Dec 14, 2012 at 10:06
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Look : http://playerio.com/

Instead of TCP or Socket, MongoDB,CouchDB, NoSQL or PlayerIO's BigDB's database solutions are better than that.

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  • \$\begingroup\$ I'm not cornerned in database solutions at the moment, but thanks \$\endgroup\$ Commented Dec 13, 2012 at 11:14

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