In C, exp() returns e raised to the power x. exp(0) is 1, and exp(1) is about 2.718.
You pass one double. The result is a positive double. Include <math.h>.
Pass the exponent. Start with this call.
exp(1.0);| Call | What it does |
|---|---|
exp(1.0) | Returns e, about 2.718. |
Include <math.h>. On GCC, link with -lm.
double exp(double x);
float expf(float x);
long double expl(long double x);| Parameter | Description |
|---|---|
| x | Required. The exponent. |
| Return Value | Description |
|---|---|
| e to the power x | A positive double. exp(0) is 1. |
Print exp(0), exp(1), and exp(2).
#include <stdio.h>
#include <math.h>
int main(void) {
printf("exp(0) = %.6f\n", exp(0.0));
printf("exp(1) = %.6f\n", exp(1.0));
printf("exp(2) = %.6f\n", exp(2.0));
return 0;
}exp.Negative exponents give values between 0 and 1—useful for decay and damping models.
#include <stdio.h>
#include <math.h>
int main(void) {
double y = exp(-1.0);
printf("exp(-1) = %.6f\n", y);
printf("1/e = %.6f\n", 1.0 / exp(1.0));
return 0;
}exp(-1) means 1/e.exp(1) matches.x, exp(x) stays > 0.x goes more negative, the result approaches 0.log()Natural log undoes exp (within floating-point error).
#include <stdio.h>
#include <math.h>
int main(void) {
double x = 2.5;
double y = exp(x);
double back = log(y);
printf("x = %.6f\n", x);
printf("exp(x) = %.6f\n", y);
printf("log(exp(x))= %.6f\n", back);
return 0;
}exp(2.5) grows to about 12.18.log(y) is the natural logarithm.log(exp(x)) returns to 2.5.log is only defined for positive arguments.Get the absolute value of a floating-point number.
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