printf problem

[Mark McLeod]

To print floats you can always use a little manipulation like:

#define ACCURACY 1000	//How many decimal places are important to you
float f;
int i, i2;   // Might need to be longs if you have a lot of significant digits
 
f = 245.56;
 
/* Print float value */ 
i = (int) f;
i2 = (int) ((f-i)*ACCURACY);
printf("f = %d.%d \r", i, i2); //e.g., f = 245.559

[Greg Ross]

Originally Posted by Mark McLeod

To print floats you can always use a little manipulation like:

#define ACCURACY 1000	//How many decimal places are important to you
float f;
int i, i2;   // Might need to be longs if you have a lot of significant digits
 
f = 245.56;
 
/* Print float value */ 
i = (int) f;
i2 = (int) ((f-i)*ACCURACY);
printf("f = %d.%d r", i, i2); //e.g., f = 245.559

You're going to want to be sure leading zeroes are printed for the fractional part. Does this printf implementation support "%03d"?

[Mark McLeod]

Originally Posted by gwross

You're going to want to be sure leading zeroes are printed for the fractional part. Does this printf implementation support "%03d"?

That's very true. I neglected that.
We should verify that the leading zeros can be forced. I'll call one of my kids later to test it on the controller at school to be sure.

[edit] That is supported (and has been tested), so the corrected printf would read:

printf("f = %d.%03d \r", i, i2);

Thanks Greg.