template<std::integral T> static inline constexpr std::string operator* (T repeat, const std::string& s) noexcept { return( s*repeat );}; A 的测试: auto main() noexcept ->int { std::cout << std::string::npos << std::endl; //constexpr -1 对应的 max_unsigned std::string s{...
#include<iostream>#include<chrono>intmain(){constexpruint32_tN=10000;volatiledoublex=3.0,y=4.0...
pow(/* floating-point-type */base, /* floating-point-type */exp) (since C++23) (constexpr since C++26) floatpow(floatbase,intexp); doublepow(doublebase,intexp); longdoublepow(longdoublebase,intexp); (2)(until C++11) floatpowf(floatbase,floatexp); ...
/* common-floating-point-type */ pow ( Arithmetic1 base, Arithmetic2 exp ); (A) (C++26 起为 constexpr) 1-4) 计算base 的exp 次幂。标准库提供所有以无 cv 限定的浮点数类型作为参数 base 和exp 的类型的 std::pow 重载。(C++23 起)A...
constexprintMOD =998244353, INV_2 = (MOD +1) /2;voidinc(int&x,inty){ x += y;if(x >= MOD) x -= MOD; }voiddec(int&x,inty){ x -= y;if(x <0) x += MOD; }intadd(intx,inty){ x += y;returnx >= MOD ? x - MOD : x; ...
(C++26 起为 constexpr) template<classArithmetic1, Arithmetic2, Arithmetic3> /*common-floating-point-type*/ hypot(Arithmetic1 x, Arithmetic2 y, Arithmetic3 z); (B)(C++17 起) 1-3)计算x和y的平方和的平方根,而不会在计算的中间阶段有过度的上溢或下溢。标准库提供所有以无 cv 限定的浮点数类型...
return std::pow(b, double(x)); } } int main() { // 常量表达式语句 constexpr double kilo = power(10.0, 3); int n = 3; // 非常量表达式,因为 n 不能在常量表达式语境中转换成右值 // 等价于 std::pow(10.0, double(n)) double mucho = power(10.0, n); std::cout << kilo << ...
std::valarray<T> pow( const std::valarray<T>& base, const std::valarray<T>& exp ); (1) template< class T > std::valarray<T> pow( const std::valarray<T>& base, const typename std::valarray<T>::value_type& vexp ); (2) template< class T > std::valarray<T> pow( ...
#pragma STDC FENV_ACCESS ON const double pi = std::acos(-1); // or std::numbers::pi since C++20 constexpr double your_sin(double x) { double sin {0}, pow {x}; for (auto fac {1LLU}, n {1ULL}; n != 20; fac *= ++n, pow *= x) if (n & 1) sin += (n & 2 ...
我正在学习c++,并且正在制作一个实时光线跟踪器。首先,我使用std::thread来分散工作,但发现每帧启动32个线程比实际需要完成的工作慢得多。然后我发现C++也通过std::async()使用线程池来处理这个问题: void trace() constexpr unsigned int thread_count = 32;[i] =std::async(std::launch:: ...