# std::frexp, std::frexpf, std::frexpl

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< cpp‎ | numeric‎ | math

Defined in header `<cmath>` |  |
| (1) |
float frexp ( float num, int* exp ); double frexp ( double num, int* exp ); long double frexp ( long double num, int* exp ); |  | (until C++23)
constexpr /* floating-point-type */ frexp ( /* floating-point-type */ num, int* exp ); |  | (since C++23)
float frexpf( float num, int* exp ); | (2) | (since C++11) (constexpr since C++23)
long double frexpl( long double num, int* exp ); | (3) | (since C++11) (constexpr since C++23)
Additional overloads (since C++11) |  |
Defined in header `<cmath>` |  |
template< class Integer > double frexp ( Integer num, int* exp ); | (A) | (constexpr since C++23)

1-3) Decomposes given floating point value num into a normalized fraction and an integral exponent of two. The library provides overloads of `std::frexp` for all cv-unqualified floating-point types as the type of the parameter num.(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
exp | - | pointer to integer value to store the exponent to

### Return value

If num is zero, returns zero and stores zero in *exp.

Otherwise (if num is not zero), if no errors occur, returns the value x in the range `(-1, -0.5], [0.5, 1)` and stores an integer value in *exp such that x×2(*exp)
== num.

If the value to be stored in *exp is outside the range of int, the behavior is unspecified.

### Error handling

This function is not subject to any errors specified in math_errhandling.

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

- If num is ±0, it is returned, unmodified, and ​0​ is stored in *exp.
- If num is ±∞, it is returned, and an unspecified value is stored in *exp.
- If num is NaN, NaN is returned, and an unspecified value is stored in *exp.
- No floating-point exceptions are raised.
- If FLT_RADIX is 2 (or a power of 2), the returned value is exact, the current rounding mode is ignored.

### Notes

On a binary system (where FLT_RADIX is 2), `std::frexp` may be implemented as

    {
        *exp = (value == 0) ? 0 : (int)(1 + std::logb(value));
        return std::scalbn(value, -(*exp));
    }

The function `std::frexp`, together with its dual, std::ldexp, can be used to manipulate the representation of a floating-point number without direct bit manipulations.

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::frexp(num, exp) has the same effect as std::frexp(static_cast<double>(num), exp).

### Example

Compares different floating-point decomposition functions:

Run this code

    #include <cmath>
    #include <iostream>
    #include <limits>

    int main()
    {
        double f = 123.45;
        std::cout << "Given the number " << f << " or " << std::hexfloat
                  << f << std::defaultfloat << " in hex,\n";

        double f3;
        double f2 = std::modf(f, &f3);
        std::cout << "modf() makes " << f3 << " + " << f2 << '\n';

        int i;
        f2 = std::frexp(f, &i);
        std::cout << "frexp() makes " << f2 << " * 2^" << i << '\n';

        i = std::ilogb(f);
        std::cout << "logb()/ilogb() make " << f / std::scalbn(1.0, i)
                  << " * " << std::numeric_limits<double>::radix
                  << "^" << std::ilogb(f) << '\n';
    }

Possible output:

    Given the number 123.45 or 0x1.edccccccccccdp+6 in hex,
    modf() makes 123 + 0.45
    frexp() makes 0.964453 * 2^7
    logb()/ilogb() make 1.92891 * 2^6

### See also

ldexpldexpfldexpl (C++11)(C++11) | multiplies a number by 2 raised to an integral power (function)
logblogbflogbl (C++11)(C++11)(C++11) | extracts exponent of the number (function)
ilogbilogbfilogbl (C++11)(C++11)(C++11) | extracts exponent of the number (function)
modfmodffmodfl (C++11)(C++11) | decomposes a number into integer and fractional parts (function)
C documentation for frexp

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