Any ideas why these 2 expressions (as far as I can see = exactly the same), give vastly different results:
I make the following change (introduce T in to the brackets...Code:Y = Y + (T/6)*(2*V + 2*P + 2*P + T*S);
Using the second line of code instead of the first gives a vastly different answer?Code:Y = Y + (1/6)*(T*2*V + T*2*P + T*2*P + T*T*S);
any ideas as to why? I am completely clueless
What are the types involved? Integer division might come into play. What are the ranges of the values involved? Maybe the result of the intermediate computations exceed the maximum value for the given type.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Well 1/6 is probably zero since you're probably using integer division. T/6 may or may not be the same thing, it depends on the type of T.
All doubles.
Oh I see. So what is the best way to perform (1/6):
?Code:static_cast<double>(1/6)
Just write (1.0 / 6.0)
That's another way of writing 0.0
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Thanks both for the help!
