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 | | 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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| 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? Here's my current copy, without (...) (24 years ago, 12-Mar-01, to lugnet.robotics.rcx.legos)
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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 (...) (24 years ago, 12-Mar-01, to lugnet.robotics.rcx.legos)
| | | | | 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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| I'm building a video game type thing. It has three touch sensors, for left, right, and fire. The LCD controls where the enemy is. I'm having some problems with my code, so here it is(I'm using kernel 0.2.4): #include "conio.h" #include "dmotor.h" (...) (24 years ago, 12-Mar-01, to lugnet.robotics.rcx.legos)
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