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  #1  
Old February 28th, 2008, 06:50 PM
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Default Re: Two battles for the price of one??

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LDiCesare said:
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vfb said:
It's the same commanders on both sides with the same names.
That doesn't mean it's not an event.
If the code goes like this:
1)Random events generate extra units.
2)Fight battles.
3)Fight battles due to random events.
4)Compute flight of surviving units.
You would see the situation you describe.
That's good thinking, but:

Random events (10) happen before movement, then other movement (13) happens, then all battles (14) caused by more than one nation (or indy) in the same province. There's no difference between "random event battles" and "movement battles". Besides, there was no random event in the province.
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Old February 29th, 2008, 04:58 AM

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Default Re: Two battles for the price of one??

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Besides, there was no random event in the province.
How can you know there's no random event in an indy province?
Depending on how (14) is implemented, I wouldn't rule out my hypothesis. Unfortunately it's almost impossible to test it.
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Old February 29th, 2008, 11:55 AM
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Default Re: Two battles for the price of one??

there could have been a random event. However, random events occur before movement, so your explanation cannot work.
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Old February 29th, 2008, 03:21 PM
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Default Re: Two battles for the price of one??

I'm pretty sure that the additional battle wasn't because of a random event. All extra battles caused by random events, barbarian raids etc show up as an independent army that attacks you but in this case it was my army that was the attacking one in both battles.
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