In C, cbrt() returns the cube root of x. cbrt(8) is 2. A negative input is allowed and gives a negative result, so cbrt(-8) is -2.
You pass one double. The result is a double with the same sign as x. Include <math.h>.
Pass any real number. Start with this call.
cbrt(8.0);| Call | What it does |
|---|---|
cbrt(8.0) | Returns 2, because 2 × 2 × 2 is 8. |
Include <math.h>. On GCC, link with -lm. Unlike sqrt(), a negative input is valid.
double cbrt(double x);
float cbrtf(float x);
long double cbrtl(long double x);| Parameter | Description |
|---|---|
| x | Required. Any real double, including negatives. |
| Return Value | Description |
|---|---|
| Cube root | A double with the same sign as x. |
Print cube roots for a few perfect cubes and zero.
#include <math.h>
#include <stdio.h>
int main(void) {
printf("cbrt(0) = %.1f\n", cbrt(0.0));
printf("cbrt(8) = %.1f\n", cbrt(8.0));
printf("cbrt(27) = %.1f\n", cbrt(27.0));
return 0;
}cbrt; link with -lm on GCC.cbrt(0) is 0.2³ = 8 and 3³ = 27.double.Cube roots of negatives are negative. Prefer cbrt over pow(x, 1.0/3.0) here.
#include <math.h>
#include <stdio.h>
int main(void) {
double x = -8.0;
double r = cbrt(x);
printf("cbrt(%.1f) = %.1f\n", x, r);
printf("check: %.1f^3 = %.1f\n", r, r * r * r);
return 0;
}cbrt stays in the reals.-2.(-2)³ = -8.pow(-8, 1.0/3.0) is often NaN on real-valued libm.For a cube, edge length is the cube root of volume.
#include <math.h>
#include <stdio.h>
int main(void) {
double volume = 64.0;
double edge = cbrt(volume);
printf("volume = %.1f\n", volume);
printf("edge = %.1f\n", edge);
return 0;
}64.4 × 4 × 4 = 64.sqrt for area → side; cbrt for volume → edge.Round a floating-point value up to the next integer.
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