How can I round a float value (such as 37.777779) to two decimal places (37.78) in C?
Use float roundf(float x)
.
"The round functions round their argument to the nearest integer value in floating-point format, rounding halfway cases away from zero, regardless of the current rounding direction." C11dr §7.12.9.5
#include
float y = roundf(x * 100.0f) / 100.0f;
Depending on your float
implementation, numbers that may appear to be half-way are not. as floating-point is typically base-2 oriented. Further, precisely rounding to the nearest 0.01
on all "half-way" cases is most challenging.
void r100(const char *s) {
float x, y;
sscanf(s, "%f", &x);
y = round(x*100.0)/100.0;
printf("%6s %.12e %.12e\n", s, x, y);
}
int main(void) {
r100("1.115");
r100("1.125");
r100("1.135");
return 0;
}
1.115 1.115000009537e+00 1.120000004768e+00
1.125 1.125000000000e+00 1.129999995232e+00
1.135 1.134999990463e+00 1.139999985695e+00
Although "1.115" is "half-way" between 1.11 and 1.12, when converted to float
, the value is 1.115000009537...
and is no longer "half-way", but closer to 1.12 and rounds to the closest float
of 1.120000004768...
"1.125" is "half-way" between 1.12 and 1.13, when converted to float
, the value is exactly 1.125
and is "half-way". It rounds toward 1.13 due to ties to even rule and rounds to the closest float
of 1.129999995232...
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float
, the value is 1.134999990463...
and is no longer "half-way", but closer to 1.13 and rounds to the closest float
of 1.129999995232...
If code used
y = roundf(x*100.0f)/100.0f;
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float
, the value is 1.134999990463...
and is no longer "half-way", but closer to 1.13 but incorrectly rounds to float
of 1.139999985695...
due to the more limited precision of float
vs. double
. This incorrect value may be viewed as correct, depending on coding goals.