Comments on: Exercise 7-8: Temperature Converter Function

By: Rhys Harris09

Rhys Harris09 — Wed, 31 Aug 2011 14:52:00 +0000

This had me stumped for ages! In France with no Internet, now in a @Cafe and have found the answer! Phew! Thanks!

By: Daniel Shiffman

Daniel Shiffman — Thu, 20 Jan 2011 14:49:47 +0000

Indeed, Processing will assume numbers are integers and you want an integer back unless you indicate a float with the decimal value (even if it is “point zero”).

By: Mary

Mary — Mon, 16 Nov 2009 14:50:56 +0000

I was tripped-up by the fractions needing to have a decimal point to work. Without the decimal point
on the 5/9 equation I was returning a 0.

I guess I should include a decimal on all floats just to be safe?

By: admin

admin — Sun, 21 Sep 2008 03:08:51 +0000

Yup, that’s right! Thanks, fixed now.

By: Aaron

Aaron — Sun, 21 Sep 2008 03:00:05 +0000

Quick question here -

In the “Answer,” shouldn’t it read:

float celsius = (farenheit – 32) * (5.0/9.0);

I could be wrong, but just wanted to check. Thanks!