C: printf a float value
CFloating PointC Problem Overview
I want to print a float value which has 2 integer digits and 6 decimal digits after the comma. If I just use printf("%f", myFloat) I'm getting a truncated value.
I don't know if this always happens in C, or it's just because I'm using C for microcontrollers (CCS to be exact), but at the reference it tells that %f get just that: a truncated float.
If my float is 44.556677, I'm printing out "44.55", only the first two decimal digits.
So the question is... how can I print my 6 digits (and just the six of them, just in case I'm having zeros after that or something)?
C Solutions
Solution 1 - C
You can do it like this:
printf("%.6f", myFloat);
6 represents the number of digits after the decimal separator.
Solution 2 - C
printf("%9.6f", myFloat) specifies a format with 9 total characters: 2 digits before the dot, the dot itself, and six digits after the dot.
Solution 3 - C
printf("%.<number>f", myFloat) //where <number> - digit after comma
Solution 4 - C
printf("%0k.yf" float_variable_name)
Here k is the total number of characters you want to get printed. k = x + 1 + y (+ 1 for the dot) and float_variable_name is the float variable that you want to get printed.
Suppose you want to print x digits before the decimal point and y digits after it. Now, if the number of digits before float_variable_name is less than x, then it will automatically prepend that many zeroes before it.
Solution 5 - C
Try these to clarify the issue of right alignment in float point printing
printf(" 4|%4.1lf\n", 8.9);
printf("04|%04.1lf\n", 8.9);
the output is
4| 8.9
04|08.9
Solution 6 - C
Use %.6f.
This will print 6 decimals.
Solution 7 - C
You need to use %2.6f instead of %f in your printf statement