Hi
Now, I do not really understand about casting, but this is not working. Could someone shine his/her light on this? Many thanks ofcourse!
unsigned char LFOendval; //between 0-128 unsigned char LFOdepth; //between 0-128 char LFOtemp= ((float)(LFOendval/128))*LFOdepth;
You need to convert your char operand to float *before* doing the arithmetic, ie:
char LFOtemp = ( (float)LFOendval / 128 ) * LFOdepth;
or even simpler:
char LFOtemp = LFOendval / 128. * LFOdepth;
Note the decimal point after 128.
So my initial solution was almost right :-)
That's nice to know! Good for my confidence ;-)
What does the point mean? Does it make the outcome a floating point number? Can I also do this:
char LFOtemp = (LFOendval / 128.) * LFOdepth;
Yes that should work.
The official way to cast is by (float)(integer_value).
But if the integer_value is a constant and not a variable, the decimal point implies a float_constant. So the math will be done in floats and then transparently cast back to whatever you are assigning to.
The rules are clear but can be confusing. So it is always safer to use the conventional cast. Just make sure you put the cast in the right place.
David.
Since the result is stored back to a char, why use floating point at all? Why not just do:
char LFOtemp = LFOendval * LFOdepth / 128;
No idea, just tried it, and does work..
Sometimes i'm just looking at it the hard way...
The simplest solution is often the best.. Thanks for pointing that out.
The simplest solution is often the best.. Thanks for pointing that out.
Not to mention that if you don't do floating point math at all, you don't reserve several hundred bytes of memory to floating point libraries and can do something else with it. Same goes with time.
Since the result is stored back to a char, why use floating point at all? Why not just do:
Code:
char LFOtemp = LFOendval * LFOdepth / 128;
Lee
