How about the following:
double d_result;
int i_result;
double remainder;
d_result = number1 / number2;
i_result = (int) d_result;
remainder = d_result - i_result;
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of jose
luis garcia
Sent: Wednesday, April 25, 2001 6:25 AM
To: Palm Developer Forum
Subject: remainder
Hello
I need to know the remainder of (number1 / number2) where number1 and
number2 there are doubles.
Is there same function for that in the codewarrior 7.0 ?
thanks....
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