送交者: Duke 于 2005-2-11, 19:32:37:

回答: casting is all right for me 由 enable 于 2005-2-11, 19:03:31:

This is what I did:

double d = 0.5;

int i = (int) 0.5 / d;

Casting there was not effective. But this works:

int i = (int) (0.5 / d);

Thanks again.

加跟贴