# std::expm1, std::expm1f, std::expm1l

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< cpp‎ | numeric‎ | math

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)
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

Run this code

    #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

exp expf expl (C++11)(C++11) | returns e raised to the given power (\({\small e^x}\)ex) (function)
exp2 exp2f exp2l (C++11)(C++11)(C++11) | returns 2 raised to the given power (\({\small 2^x}\)2x) (function)
log1p log1pf log1pl (C++11)(C++11)(C++11) | natural logarithm (to base e) of 1 plus the given number (\({\small \ln{(1+x)} }\)ln(1+x)) (function)

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