6.11.117 log1pl Function

Calculates the natural logarithm of a long double precision floating-point value plus 1.

Include

<math.h>

Prototype

long double log1pl(long double x);

Argument

x

any positive value for which to return the log

Return Value

Returns the natural (base-$e$) logarithm of x+1 or NaN if a domain error occurs. The result of log1p(x) is generally expected to be more accurate than that of log(x+1) when the magnitude of x is small.

Remarks

If x is less than -1, a domain error will occur and errno will be set to EDOM.

Example

See the notes at the beginning of this chapter or section for information on using printf() or scanf() (and other functions reading and writing the stdin or stdout streams) in the example code.

#include <math.h>
#include <stdio.h>
#include <errno.h>

int main(void)
{
  double x, y;

  errno = 0;
  x = 2.0;
  y = log1pl(x);
  if(errno)
    perror("Error");
  printf("The result of log1pl(%f) is %f\n", x, y);

  errno = 0;
  x = 0.0;
  y = log1pl(x);
  if(errno)
    perror("Error");
  printf("The result of log1pl(%f) is %f\n", x, y);

  errno = 0;
  x = -2.0;
  y = log1pl(x);
  if(errno)
    perror("Error");
  printf("The result of log1pl(%f) is %f\n", x, y);
}

Example Output

The result of log1pl(2.000000) is 1.098612
The result of log1pl(0.000000) is 0.000000
Error: domain error
The result of log1pl(-2.000000) is nan