expm1, expm1f, expm1l

Defined in header <math.h>
float       expm1f( float arg );
(1) (since C99)
double      expm1( double arg );
(2) (since C99)
long double expm1l( long double arg );
(3) (since C99)
Defined in header <tgmath.h>
#define expm1( arg )
(4) (since C99)
1-3) Computes the e (Euler's number, 2.7182818) raised to the given power arg, minus 1.0. This function is more accurate than the expression exp(arg)-1.0 if arg is close to zero.
4) Type-generic macro: If arg has type long double, expm1l is called. Otherwise, if arg has integer type or the type double, expm1 is called. Otherwise, expm1f is called.


arg - floating point value

Return value

If no errors occur earg
-1 is returned.

If a range error due to overflow occurs, +HUGE_VAL, +HUGE_VALF, or +HUGE_VALL is returned.

If a range error occurs due to underflow, the correct result (after rounding) is returned.

Error handling

Errors are reported as specified in math_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

  • If the argument is ±0, it is returned, unmodified
  • If the argument is -∞, -1 is returned
  • If the argument is +∞, +∞ is returned
  • If the argument is NaN, NaN is returned


The functions expm1 and log1p are useful for financial calculations, for example, when calculating small daily interest rates: (1+x)n
-1 can be expressed as expm1(n * log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.

For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg.


#include <stdio.h>
#include <math.h>
#include <float.h>
#include <errno.h>
#include <fenv.h>
int main(void)
    printf("expm1(1) = %f\n", expm1(1));
    printf("Interest earned in 2 days on $100, compounded daily at 1%%\n"
           " on a 30/360 calendar = %f\n",
    printf("exp(1e-16)-1 = %g, but expm1(1e-16) = %g\n",
           exp(1e-16)-1, expm1(1e-16));
    // special values
    printf("expm1(-0) = %f\n", expm1(-0.0));
    printf("expm1(-Inf) = %f\n", expm1(-INFINITY));
    //error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);
    printf("expm1(710) = %f\n", expm1(710));
    if(errno == ERANGE) perror("    errno == ERANGE");
    if(fetestexcept(FE_OVERFLOW)) puts("    FE_OVERFLOW raised");

Possible output:

expm1(1) = 1.718282
Interest earned in 2 days on $100, compounded daily at 1%
 on a 30/360 calendar = 0.005556
exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16
expm1(-0) = -0.000000
expm1(-Inf) = -1.000000
expm1(710) = inf
    errno == ERANGE: Result too large
    FE_OVERFLOW raised


  • C17 standard (ISO/IEC 9899:2018):
    • The expm1 functions (p: 177)
    • 7.25 Type-generic math <tgmath.h> (p: 272-273)
    • F.10.3.3 The expm1 functions (p: 379)
  • C11 standard (ISO/IEC 9899:2011):
    • The expm1 functions (p: 243)
    • 7.25 Type-generic math <tgmath.h> (p: 373-375)
    • F.10.3.3 The expm1 functions (p: 521)
  • C99 standard (ISO/IEC 9899:1999):
    • The expm1 functions (p: 223-224)
    • 7.22 Type-generic math <tgmath.h> (p: 335-337)
    • F.9.3.3 The expm1 functions (p: 458)

See also

computes e raised to the given power (\({\small e^x}\)ex)
computes 2 raised to the given power (\({\small 2^x}\)2x)
computes natural (base-e) logarithm of 1 plus the given number (\({\small \ln{(1+x)} }\)ln(1+x))
C++ documentation for expm1

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