我搜索了一圈,但找不到 C++ 的 trunc
函数。我知道可以这样做:
int main()
{
double a = 12.566789;
cout << setprecision(2) << fixed << (int)(a * 100) / 100.0 << endl;
return 0;
}
但我不确定这是否是最佳方法。谢谢。
trunc
函数,则可以基于C89规范中指定的floor
和ceil
函数轻松实现一个。double trunc(double d){ return (d>0) ? floor(d) : ceil(d) ; }
trunc
是在 <cmath>
中的:
#include <iostream>
#include <cmath>
int main() {
std::cout << trunc(3.141516) << std::endl;
}
C语言中有一个截断函数,你也可以在C++中使用它。
trunc(a*100)/100
请记住,您仍然需要指定格式请求,因为浮点数无法精确表示所有实数,如果您不告诉输出代码您想要的精度,您可能会得到如 12.5600000001
或 12.55999999
的输出。
简而言之:
使用以下内容进行输出:
cout << setprecision(2) << fixed << a<< endl;
如果在数学计算过程中需要截断结果,则可以使用以下方法:
trunc(a*100)/100
double
时,您需要使用 trunc
而不是 truncl
(后者适用于 long double
)。 - Stephen Canonsetprecision(2)
会四舍五入值,而不是截断。 - Alireza Nooritrunc
在C99标准中被添加,因此在某些不带现代C库的旧版C++编译器中不存在。 C++11最终将其添加到C标准中,因此希望那些供应商在未来几年内终于现代化其数学库。 - Stephen Canondouble a = 12.566789;
double b = trunc(a * 100) / 100.0;
如果你正在使用一个没有实现trunc
的古老的C或C++库,可以使用boost::math::trunc
。
我开发了一个非常快速的截断函数:
double ftrunc( double d )
{
static_assert(sizeof(double) == sizeof(uint64_t), "sizeof(double) not equal to sizeof(uint64_t)");
static_assert(numeric_limits<double>::is_iec559, "double must be IEEE-754");
// assume size_t is our CPU's native register-width
static_assert(sizeof(size_t) == sizeof(uint64_t) || sizeof(size_t) == sizeof(uint32_t), "register-width must be 32 or 64 bit");
if constexpr( sizeof(size_t) == sizeof(uint64_t) )
// we have 64 bit registers
{
unsigned const MANTISSA_BITS = 52,
EXP_BIAS = 0x3FF,
INF_NAN_BASE = 0x7FF;
uint64_t const EXP_MASK = (uint64_t)0x7FF << MANTISSA_BITS,
SIGN_MASK = (uint64_t)0x800 << MANTISSA_BITS ,
MIN_INTEGRAL_DIGITS_EXP = (uint64_t) EXP_BIAS << MANTISSA_BITS,
MIN_INTEGRAL_ONLY_EXP = (uint64_t)(EXP_BIAS + MANTISSA_BITS) << MANTISSA_BITS,
INF_NAN_EXP = (uint64_t)INF_NAN_BASE << MANTISSA_BITS,
NEG_MANTISSA_MASK = 0x000FFFFFFFFFFFFFu;
union
{
double du;
uint64_t dx;
};
du = d;
uint64_t exp = dx & EXP_MASK;
if( exp >= MIN_INTEGRAL_DIGITS_EXP )
// value has integral digits
if( exp < MIN_INTEGRAL_ONLY_EXP )
{
// there are fraction-digits to mask out, mask them
unsigned shift = (unsigned)(exp >> MANTISSA_BITS) - EXP_BIAS;
dx &= ~(NEG_MANTISSA_MASK >> shift);
return du;
}
else
if( exp < INF_NAN_EXP )
// value is integral
return du;
else
// infinite, NaN, SNaN
// raise exception on SNaN if necessary
return du + du;
else
{
// below +/-1.0
// return +/-0.0
dx &= SIGN_MASK;
return du;
}
}
else if constexpr( sizeof(size_t) == sizeof(uint32_t) )
// we have 32 bit registers
{
unsigned const MANTISSA_BITS = 52,
HI_MANTISSA_BITS = 20,
EXP_BIAS = 0x3FF,
INF_NAN_BASE = 0x7FF;
uint32_t const EXP_MASK = (uint32_t)0x7FFu << HI_MANTISSA_BITS,
SIGN_MASK = (uint32_t)0x800u << HI_MANTISSA_BITS,
MIN_INTEGRAL_DIGITS_EXP = (uint32_t) EXP_BIAS << HI_MANTISSA_BITS,
MAX_INTEGRAL32_EXP = (uint32_t)(EXP_BIAS + HI_MANTISSA_BITS) << HI_MANTISSA_BITS,
MIN_INTEGRAL_ONLY_EXP = (uint32_t)(EXP_BIAS + MANTISSA_BITS) << HI_MANTISSA_BITS,
INF_NAN_EXP = (uint32_t)INF_NAN_BASE << HI_MANTISSA_BITS,
NEG_HI_MANTISSA_MASK = 0x000FFFFFu,
NEG_LO_MANTISSA_MASK = 0xFFFFFFFFu;
union
{
double du;
struct
{
uint32_t dxLo;
uint32_t dxHi;
};
};
du = d;
uint32_t exp = dxHi & EXP_MASK;
if( exp >= MIN_INTEGRAL_DIGITS_EXP )
// value has integral digits
if( exp < MIN_INTEGRAL_ONLY_EXP )
// there are fraction-digits to mask out
if( exp <= MAX_INTEGRAL32_EXP )
{
// the fraction digits are in the upper dword, mask them and zero the lower dword
unsigned shift = (unsigned)(exp >> HI_MANTISSA_BITS) - EXP_BIAS;
dxHi &= ~(NEG_HI_MANTISSA_MASK >> shift);
dxLo = 0;
return du;
}
else
{
// the fraction digits are in the lower dword, mask them
unsigned shift = (unsigned)(exp >> HI_MANTISSA_BITS) - EXP_BIAS - HI_MANTISSA_BITS;
dxLo &= ~(NEG_LO_MANTISSA_MASK >> shift);
return du;
}
else
if( exp < INF_NAN_EXP )
// value is integral
return du;
else
// infinite, NaN, SNaN
// raise exception on SNaN if necessary
return du + du;
else
{
// below +/-1.0
// return +/-0.0
dxHi &= SIGN_MASK;
dxLo = 0;
return du;
}
}
}
它比大多数实现都要快。在我的Ryzen 7 1800X上,值>= 2^0且<= 2^54的平均执行时间为12个时钟周期。
使用cmath中的ceil
或floor
trunc()
是另一种方法。你正在寻找一个截断(trunc)函数,而不是一个大的脚储物柜(trunk)函数。 - David Hammen