std::bit_cast

Defined in header `<bit>`
template< class To, class From > constexpr To bit_cast( const From& from ) noexcept;		(since C++20)

Obtain a value of type To by reinterpreting the object representation of From. Every bit in the value representation of the returned To object is equal to the corresponding bit in the object representation of from. The values of padding bits in the returned To object are unspecified.

If there is no value of type To corresponding to the value representation produced, the behavior is undefined. If there are multiple such values, which value is produced is unspecified.

A bit in the value representation of the result is indeterminate if it

does not correspond to a bit in the value representation of From (i.e. it corresponds to a padding bit), or
corresponds to a bit of an object that is not within its lifetime, or
has an indeterminate value.

For each bit in the value representation of the result that is indeterminate, the smallest object containing that bit has an indeterminate value; the behavior is undefined unless that object is of unsigned char or std::byte type. The result does not otherwise contain any indeterminate values.

This overload participates in overload resolution only if sizeof(To) == sizeof(From) and both To and From are TriviallyCopyable types.

This function template is constexpr if and only if each of To, From and the types of all subobjects of To and From:

is not a union type;
is not a pointer type;
is not a pointer to member type;
is not a volatile-qualified type; and
has no non-static data member of reference type.

Parameters

from

-

the source of bits for the return value

Return value

An object of type To whose value representation is as described above.

Possible implementation

To implement std::bit_cast, ignoring the fact that it's constexpr, std::memcpy can be used, when it is needed, to interpret the object representation as one of another type:

template<class To, class From>
std::enable_if_t<
    sizeof(To) == sizeof(From) &&
    std::is_trivially_copyable_v<From> &&
    std::is_trivially_copyable_v<To>,
    To>
// constexpr support needs compiler magic
bit_cast(const From& src) noexcept
{
    static_assert(std::is_trivially_constructible_v<To>,
        "This implementation additionally requires "
        "destination type to be trivially constructible");
 
    To dst;
    std::memcpy(&dst, &src, sizeof(To));
    return dst;
}

Notes

reinterpret_cast (or equivalent explicit cast) between pointer or reference types shall not be used to reinterpret object representation in most cases because of the type aliasing rule.

Feature-test macro	Value	Std	Comment
`__cpp_lib_bit_cast`	201806L	(C++20)	`std::bit_cast`

Example

Run this code

#include <bit>
#include <cstdint>
#include <iostream>
 
constexpr double f64v = 19880124.0; 
constexpr auto u64v = std::bit_cast<std::uint64_t>(f64v);
static_assert(std::bit_cast<double>(u64v) == f64v); // round-trip
 
constexpr std::uint64_t u64v2 = 0x3fe9000000000000ull;
constexpr auto f64v2 = std::bit_cast<double>(u64v2);
static_assert(std::bit_cast<std::uint64_t>(f64v2) == u64v2); // round-trip
 
int main()
{
    std::cout
        << "std::bit_cast<std::uint64_t>(" << std::fixed << f64v << ") == 0x"
        << std::hex << u64v << '\n'
        << "std::bit_cast<double>(0x" << std::hex << u64v2 << ") == "
        << std::fixed << f64v2 << '\n';
}

Possible output:

std::bit_cast<std::uint64_t>(19880124.000000) == 0x4172f58bc0000000
std::bit_cast<double>(0x3fe9000000000000) == 0.781250

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR	Applied to	Behavior as published	Correct behavior
CWG 2482	C++20	it was unspecified whether UB would occur when involving indeterminate bits	specified

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