If you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:
#include <math.h>
float val = 37.777779;
float rounded_down = floorf(val * 100) / 100; /* Result: 37.77 */
float nearest = roundf(val * 100) / 100; /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100; /* Result: 37.78 */
Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.
As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.
For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.
Answer from Dale Hagglund on Stack OverflowIf you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:
#include <math.h>
float val = 37.777779;
float rounded_down = floorf(val * 100) / 100; /* Result: 37.77 */
float nearest = roundf(val * 100) / 100; /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100; /* Result: 37.78 */
Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.
As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.
For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.
Using %.2f in printf. It only print 2 decimal points.
Example:
printf("%.2f", 37.777779);
Output:
37.77
CCS :: View topic - Float Values in Printf() upto 2 decimal values
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i'd suggest shorter and faster approach:
printf("%.2f", ((signed long)(fVal * 100) * 0.01f));
this way you won't overflow int, plus multiplication by 100 shouldn't influence the significand/mantissa itself, because the only thing that really is changing is exponent.
The only easy way to do this is to use snprintf to print to a buffer that's long enough to hold the entire, exact value, then truncate it as a string. Something like:
char buf[2*(DBL_MANT_DIG + DBL_MAX_EXP)];
snprintf(buf, sizeof buf, "%.*f", (int)sizeof buf, x);
char *p = strchr(buf, '.'); // beware locale-specific radix char, though!
p[2+1] = 0;
puts(buf);
This is a very simple question.
For example: Instead of "5.600000000" appearing on the screen, "5.600" appears instead.
I think there was a way but I don't remember it.
I’m new to c and I would appreciate any kind of help
I have a float variable that only goes to 2 decimal points and when I cast it as a string to a text field it would only show 1 or none if the rest are 0. For example 10.90 would be 10.9 and 1.00 would be just 1.
Edit: sorry I said it in a very confusing way. This is what currently happens.
var total = 10.90
text = str(total)
the text field would be 10.9
I’m looking for a way for it to show 10.90 I want it to always be on 2 decimal places. Hope this clarifies things.
Thanks for the help.