C: printf a float value

Question

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)?

Answer 1

You can do it like this:

printf("%.6f", myFloat);

6 represents the number of digits after the decimal separator.

Answer 2

printf("%9.6f", myFloat)指定了一种共有 9 个字符的格式：点前 2 位数字、点本身和点后 6 位数字。

Answer 3

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.

Answer 4

printf("%.<number>f", myFloat) //where <number> - digit after comma

http://www.cplusplus.com/reference/clibrary/cstdio/printf/

Answer 5

Use %.6f . This will print 6 decimals.

Answer 6

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

Answer 7

您需要在 printf 语句中使用%2.6f而不是%f