Home Original page

std::expm1, std::expm1f, std::expm1l - cppreference.com

From cppreference.com

Defined in header <cmath>

(1)

float expm1 ( float num ); double expm1 ( double num ); long double expm1 ( long double num );

 (until C++23)

/*floating-point-type*/ expm1 ( /*floating-point-type*/ num );

 (since C++23)
(constexpr since C++26)

float expm1f( float num );

 (2) (since C++11)
(constexpr since C++26)

long double expm1l( long double num );

 (3) (since C++11)
(constexpr since C++26)

SIMD overload (since C++26)

Defined in header <simd>

template< /*math-floating-point*/ V > constexpr /*deduced-simd-t*/<V> expm1 ( const V& v_num );

 (S) (since C++26)

Additional overloads (since C++11)

Defined in header <cmath>

template< class Integer > double expm1 ( Integer num );

 (A) (constexpr since C++26)

1-3) Computes the e (Euler's number, 2.7182818...) raised to the given power num, minus 1.0. This function is more accurate than the expression std::exp(num) - 1.0 if num is close to zero. The library provides overloads of std::expm1 for all cv-unqualified floating-point types as the type of the parameter.(since C++23)

A) Additional overloads are provided for all integer types, which are treated as double.

 (since C++11)

Parameters

num - floating-point or integer value

Return value

If no errors occur enum
-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.

Notes

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

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

The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their argument num of integer type, std::expm1(num) has the same effect as std::expm1(static_cast<double>(num)).

Example

#include <cerrno>
#include <cfenv>
#include <cmath>
#include <cstring>
#include <iostream>
// #pragma STDC FENV_ACCESS ON

int main()
{
    std::cout << "expm1(1) = " << std::expm1(1) << '\n'
              << "Interest earned in 2 days on $100, compounded daily at 1%\n"
              << "    on a 30/360 calendar = "
              << 100 * std::expm1(2 * std::log1p(0.01 / 360)) << '\n'
              << "exp(1e-16)-1 = " << std::exp(1e-16) - 1
              << ", but expm1(1e-16) = " << std::expm1(1e-16) << '\n';
    
    // special values
    std::cout << "expm1(-0) = " << std::expm1(-0.0) << '\n'
              << "expm1(-Inf) = " << std::expm1(-INFINITY) << '\n';
    
    // error handling
    errno = 0;
    std::feclearexcept(FE_ALL_EXCEPT);
    
    std::cout << "expm1(710) = " << std::expm1(710) << '\n';
    
    if (errno == ERANGE)
        std::cout << "    errno == ERANGE: " << std::strerror(errno) << '\n';
    if (std::fetestexcept(FE_OVERFLOW))
        std::cout << "    FE_OVERFLOW raised\n";
}

Possible output: 

expm1(1) = 1.71828
Interest earned in 2 days on $100, compounded daily at 1%
    on a 30/360 calendar = 0.00555563
exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16
expm1(-0) = -0
expm1(-Inf) = -1
expm1(710) = inf
    errno == ERANGE: Result too large
    FE_OVERFLOW raised

See also