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Showing content from https://en.cppreference.com/w/cpp/language/../numeric/math/frexp.html below:

std::frexp, std::frexpf, std::frexpl - cppreference.com

(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)

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) [edit] Parameters num - floating-point or integer value exp - pointer to integer value to store the exponent to [edit] 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.

[edit] Error handling

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

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

[edit] Notes

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

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).

[edit] Example

Compares different floating-point decomposition functions:

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
[edit] See also

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