arithmetic precedence head scratcher (noob) C++ VS2005Pro

[jbe3d]

Here is a snippet of code that returns 0 (c = any number)
d = 5 / 9 * (c-32);

here it works as I expect it should, resulting in the expected numeric value
d = (c-32) * 5 / 9;

I'm spending time reading though MSDN but have only come up with an indication that that (c-32) in the first example might be an empty expression.

Thanks in advance for helping me to understand why the expression gives two different results when the two sides are swapped.

[dtimes6]

(5/9) * (c - 32) != ((c-32) * 5)/9 if c and d are types like int or char.

[Banfa]

5 and 9 are integers and / * have the same precedence so it just does the calculation in order

d = 5 / 9 * (c-32);

5 / 9 = 0 (under integer division)

0 * (c-32) = 0 for any c

However here

d = (c-32) * 5 / 9;

the * is first so first it does assuming c = -40

(c-32) = -40 - 32 = -72

-72 * 5 = -360

-360 / 9 = -40



If is normal when doing integer arithmatic to perform multiplication before divison for this reason, it preserves the result and gives better a better result consider the difference between

4321 / 7 * 10 = 6170 (under integer arithmatic)
and
4321 * 10 / 7 = 6172 (under integer arithmatic)

The only thing you have to watch out for is that you don't exceed the limits of you variable type (int in this case) when performing the multiplication.