std::bit_width - cppreference.com (original) (raw)

| Defined in header | | | | -------------------------------------------------------------- | | ------------- | | template< class T > constexpr int bit_width( T x ) noexcept; | | (since C++20) |

If x is not zero, calculates the number of bits needed to store the value x, that is, \(1 + \lfloor \log_2(x) \rfloor\)1 + floor(log2(x)). If x is zero, returns zero.

This overload participates in overload resolution only if T is an unsigned integer type (that is, unsigned char, unsigned short, unsigned int, unsigned long, unsigned long long, or an extended unsigned integer type).

[edit] Parameters

[edit] Return value

Zero if x is zero; otherwise, one plus the base-2 logarithm of x, with any fractional part discarded.

[edit] Notes

This function is equivalent to return std::numeric_limits<T>::digits - std::countl_zero(x);.

[edit] Example

#include #include #include   int main() { for (unsigned x{}; x != 010; ++x) std::cout << "bit_width( " << std::bitset<4>{x} << " ) = " << std::bit_width(x) << '\n'; }

Output:

bit_width( 0000 ) = 0 bit_width( 0001 ) = 1 bit_width( 0010 ) = 2 bit_width( 0011 ) = 2 bit_width( 0100 ) = 3 bit_width( 0101 ) = 3 bit_width( 0110 ) = 3 bit_width( 0111 ) = 3

[edit] Defect reports

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

[edit] See also

| | counts the number of consecutive ​0​ bits, starting from the most significant bit (function template) [edit] | | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |