AVR Freaks

You need to beware of rounding:

 

 Serial.print(DutyCycle, 2);

just to point out that .print() is known as an overloaded function. It may have one name but it has:

Print.h:    size_t print(const __FlashStringHelper *);
Print.h:    size_t print(const String &);
Print.h:    size_t print(const char[]);
Print.h:    size_t print(char);
Print.h:    size_t print(unsigned char, int = DEC);
Print.h:    size_t print(int, int = DEC);
Print.h:    size_t print(unsigned int, int = DEC);
Print.h:    size_t print(long, int = DEC);
Print.h:    size_t print(unsigned long, int = DEC);
Print.h:    size_t print(double, int = 2);
Print.h:    size_t print(const Printable&);

eleven different implementations. If you pass it an unsigned char for example it will use the:

Print.h:    size_t print(unsigned char, int = DEC);

version. In fact that even shows another feature of C++ which is "default parameters". So you can write:

unsigned char n = 123;

Serial.print(n);

and it will print "123" (so there is a kind of implied itoa() here or maybe a sprintf("%u") ) The value is printed in decimal because of the second (not given in this case) parameter defaulting to DEC. But you could have used:

Serial.print(n, HEX);

and the 2nd parameter you provide will over-ride the "DEC" that it would be taken to be by default and so print the value in hex ( so "7B").

 

Anyway the point is that Serial.print() will adapt to pretty much anything you throw at it and have a go at printing it in the right format. So unlike a traditional C function where you might write:

void print(unsigned char n) {
    send_to_output(n);
}

and then if you use:

print(3.141592);

this would print "3" because there only is one version of print() and because the input interface says "I take an unsigned char" then when you pass it 3.141592 C knows you can't fit a double into an unsigned char so it does an implied conversion for you here. But that is not the way C++ works when you have an overloaded function like Serial.print(). So if you

Serial.print(3.141592);

expect to see "3.141592" at the output as it looks through the versions it has and says to itself the parameter is a double so I will use this one this time:

Print.h:    size_t print(double, int = 2);

(oh and this one has a default parameter too but this time I think it's being used to say the number of decimal places).

 

Therefore to overcome this additional cleverness you need to cast:

Serial.print((unsigned char)3.141592);

really would print "3". because it takes the double and converts it to unsigned char before passing the value to the function and that means it will pick the "unsigned char" version.  

 

In fact "(unsigned char)" in this is " C style cast". C++ has several different ways to cast (most concerned with the handling of "const" in fact). So the one you would probably use in C++ is:

Serial.print(static_cast<unsigned char>(3.141592));

Of course:

double f = 3.141592;
unsigned char n;

n = f;
Serial.print(n);

would also work as the assignment does an implicit case from double to unsigned char.

 

(BTW note I have been using the word "double". That is because 3.141592 is a double value. It is only a "float" value if you express it as 3.141592F )

Do you mean a rounding error? During conversion C simply truncates So 3.01 to 3.99  will all be rounded down to 3. If you want it to round up from 3.5 upwards add 0.5 to it. So if, for example the value is 3.71 then 3.71 + 0.5 is 4.21 then usual truncation will make this just 4 when converted to integer. If, on the other hand it had been 3.49 then 3.49 + 0.5 is 3.99 and that still becomes just 3. So anything from n.50 to n.99 will round up to n+1 but anything from n.00 to n.49 will be just n.

converted = (DutyCycle + 0.5);