In C, floor() rounds x down toward negative infinity. floor(2.9) is 2, and floor(-2.1) is -3.
You pass one double. The result is still a double, not an int. Include <math.h>.
Pass the number you want to round down. Start with this call.
floor(2.9);| Call | What it does |
|---|---|
floor(2.9) | Rounds down to 2. |
Include <math.h>. On GCC, link with -lm. This is not the same as cutting off the fraction. For that, use trunc().
double floor(double x);
float floorf(float x);
long double floorl(long double x);| Parameter | Description |
|---|---|
| x | Required. The double to round toward -infinity. |
| Return Value | Description |
|---|---|
| double | The largest integer value that is not greater than x, stored as a double. |
Round a positive non-integer down to the previous whole number.
#include <stdio.h>
#include <math.h>
int main(void) {
printf("%.1f\n", floor(5.78));
printf("%.1f\n", floor(5.0));
return 0;
}floor.%.1f shows one decimal place.Toward -infinity means negatives with a fraction become more negative—not closer to zero.
#include <stdio.h>
#include <math.h>
int main(void) {
printf("floor(-2.3) = %.1f\n", floor(-2.3));
printf("ceil(-2.3) = %.1f\n", ceil(-2.3));
return 0;
}ceil(-2.3) is -2.trunc.Count complete groups when a partial group does not count.
#include <stdio.h>
#include <math.h>
int main(void) {
int items = 25;
int batch = 10;
int full = (int)floor(items / (double)batch);
printf("full batches = %d\n", full);
return 0;
}25 / (double)10 is 2.5.floor(2.5) is 2.0—only complete batches.ceil; leftover ignored uses floor.Get the floating-point remainder after dividing two doubles.
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