Subject:
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Re: video game type thing...
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Newsgroups:
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lugnet.robotics.rcx.legos
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Date:
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Mon, 12 Mar 2001 20:11:39 GMT
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Viewed:
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1422 times
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Thanks for the response! Now I'm almost ready to test it. I just need to
know if, when you have a call void function, and you then use the int main()
to close the program with return 0;, will the void keep running?
Thanks
Nick Howell
In lugnet.robotics.rcx.legos, Michael Ash writes:
> On Mon, 12 Mar 2001, Nick Howell wrote:
>
> > if(enemy = 1)
> > {
> > fight;
> > monitor;
> > }
>
> I think you mean if (enemy == 1).
>
> Also, the identifier "fight" resolves to the address of the fight
> function. Doing "fight;" is equivalent to doing "1;". It's legal, but it
> doesn't accomplish anything. You want to do "fight();" and "monitor();".
>
> You'll probably have trouble with the compiler not knowing what you're
> talking about at this point. You need to either provide prototypes for
> your function (put "void fight(void);" and the like at the beginning of
> your file), or have the actual function itself in the file before you call
> it.
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Message has 1 Reply:
 | | Re: video game type thing...
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| (...) No, it will not. Functions don't work like that, in general. If you want to call a function and have it run independently of the calling function, you must use execi() to start it. (24 years ago, 12-Mar-01, to lugnet.robotics.rcx.legos)
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Message is in Reply To:
| | Re: video game type thing...
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| (...) I think you mean if (enemy == 1). Also, the identifier "fight" resolves to the address of the fight function. Doing "fight;" is equivalent to doing "1;". It's legal, but it doesn't accomplish anything. You want to do "fight();" and (...) (24 years ago, 12-Mar-01, to lugnet.robotics.rcx.legos)
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