ansaurus

Question

Answer 1

+4 A:

Convert one of them to a double first. This form works in many languages:

 real_result = (int_numerator + 0.0) / int_denominator

Mark Ransom 2009-03-19 04:12:02

Answer 2

+5 A:

cast the integers to doubles.

Stephen Wrighton 2009-03-19 04:12:47

Answer 3

+20 A:

You want to cast the numbers:

    double num3 = (double)num1/(double)num2;

Note: If any of the arguments in C# are doubles, a double divide is used which results in a double. So, the following would work too:

    double num3 = (double)num1/num2;

For more information see:

NoahD 2009-03-19 04:14:16

Don't know if this is the same in C#, but C only requires you to cast the first - it'll automatically make double/int a double.

paxdiablo 2009-03-19 04:34:09

@Pax, If any of the args in C or C# are a double, a double divide is used (resulting in a double).

strager 2009-03-19 05:18:57

How can I divide two integers to get a double?