我想写一个函数，返回最接近的2的次幂数。例如，如果我的输入是789，输出应该是1024。有没有办法不使用任何循环，而只是使用一些位运算符?
当前回答
c++ 14 clp2的constexpr版本
#include <iostream>
#include <type_traits>
// Closest least power of 2 minus 1. Returns 0 if n = 0.
template <typename UInt, std::enable_if_t<std::is_unsigned<UInt>::value,int> = 0>
constexpr UInt clp2m1(UInt n, unsigned i = 1) noexcept
{ return i < sizeof(UInt) * 8 ? clp2m1(UInt(n | (n >> i)),i << 1) : n; }
/// Closest least power of 2 minus 1. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value && std::is_signed<Int>::value,int> = 0>
constexpr auto clp2m1(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n <= 0 ? 0 : n)); }
/// Closest least power of 2. Returns 2^N: 2^(N-1) < n <= 2^N. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
constexpr auto clp2(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n-1)) + 1; }
/// Next power of 2. Returns 2^N: 2^(N-1) <= n < 2^N. Returns 1 if n = 0. Returns 0 if n < 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
constexpr auto np2(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n)) + 1; }
template <typename T>
void test(T v) { std::cout << clp2(v) << std::endl; }
int main()
{
test(-5); // 0
test(0); // 0
test(8); // 8
test(31); // 32
test(33); // 64
test(789); // 1024
test(char(260)); // 4
test(unsigned(-1) - 1); // 0
test<long long>(unsigned(-1) - 1); // 4294967296
return 0;
}
其他回答
这是我用来让它成为一个常数表达式的，如果输入是一个常数表达式的话。
#define uptopow2_0(v) ((v) - 1)
#define uptopow2_1(v) (uptopow2_0(v) | uptopow2_0(v) >> 1)
#define uptopow2_2(v) (uptopow2_1(v) | uptopow2_1(v) >> 2)
#define uptopow2_3(v) (uptopow2_2(v) | uptopow2_2(v) >> 4)
#define uptopow2_4(v) (uptopow2_3(v) | uptopow2_3(v) >> 8)
#define uptopow2_5(v) (uptopow2_4(v) | uptopow2_4(v) >> 16)
#define uptopow2(v) (uptopow2_5(v) + 1) /* this is the one programmer uses */
例如，这样的表达式:
uptopow2(sizeof (struct foo))
会很好地化简为常数。
我试着找到最接近2的次幂，然后写出这个函数。希望它能帮助你。只要用最近的小数乘以2，就能得到2的最近上次方
int nearest_upper_power(int number){
int temp=number;
while((number&(number-1))!=0){
temp<<=1;
number&=temp;
}
//Here number is closest lower power
number*=2;
return number;
}
如果您想要单行模板。在这里
int nxt_po2(int n) { return 1 + (n|=(n|=(n|=(n|=(n|=(n-=1)>>1)>>2)>>4)>>8)>>16); }
or
int nxt_po2(int n) { return 1 + (n|=(n|=(n|=(n|=(n|=(n-=1)>>(1<<0))>>(1<<1))>>(1<<2))>>(1<<3))>>(1<<4)); }
许多处理器架构都支持log以2为底或非常类似的操作——计数前导零。许多编译器都有针对它的内在特性。参见https://en.wikipedia.org/wiki/Find_first_set
c++ 14 clp2的constexpr版本
#include <iostream>
#include <type_traits>
// Closest least power of 2 minus 1. Returns 0 if n = 0.
template <typename UInt, std::enable_if_t<std::is_unsigned<UInt>::value,int> = 0>
constexpr UInt clp2m1(UInt n, unsigned i = 1) noexcept
{ return i < sizeof(UInt) * 8 ? clp2m1(UInt(n | (n >> i)),i << 1) : n; }
/// Closest least power of 2 minus 1. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value && std::is_signed<Int>::value,int> = 0>
constexpr auto clp2m1(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n <= 0 ? 0 : n)); }
/// Closest least power of 2. Returns 2^N: 2^(N-1) < n <= 2^N. Returns 0 if n <= 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
constexpr auto clp2(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n-1)) + 1; }
/// Next power of 2. Returns 2^N: 2^(N-1) <= n < 2^N. Returns 1 if n = 0. Returns 0 if n < 0.
template <typename Int, std::enable_if_t<std::is_integral<Int>::value,int> = 0>
constexpr auto np2(Int n) noexcept
{ return clp2m1(std::make_unsigned_t<Int>(n)) + 1; }
template <typename T>
void test(T v) { std::cout << clp2(v) << std::endl; }
int main()
{
test(-5); // 0
test(0); // 0
test(8); // 8
test(31); // 32
test(33); // 64
test(789); // 1024
test(char(260)); // 4
test(unsigned(-1) - 1); // 0
test<long long>(unsigned(-1) - 1); // 4294967296
return 0;
}
