如何在C ++中编写良好的round_double函数？
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我正在尝试编写良好的round_double函数，它将以指定的精度四舍五入：
1。
double round_double(double num, int prec)
{
for (int i = 0; i < abs(prec); ++i)
if(prec > 0)
num *= 10.0;
else
num /= 10.0;
double result = (long long)floor(num + 0.5);
for (int i = 0; i < abs(prec); ++i)
if(prec > 0)
result /= 10.0;
else
result *= 10.0;
return result;
}
double round_double(double num, int prec)
{
double tmp = pow(10.0, prec);
double result = (long long)floor(num * tmp + 0.5);
result /= tmp;
return result;
}
cout
int prec = 18;
double num = 10.123456789987654321;
cout << setiosflags(ios::showpoint | ios::fixed)
<< setprecision(prec) << \"round_double(\" << num << \", \"
<< prec << \") = \" << round_double(num, prec) << endl;
round_double(10.123456789987655000, 18) = -9.223372036854776500
没有找到相关结果
已邀请:
3 个回复
旅牢斯讲
就足够了。double round_double(double num, int prec) { double tmp = pow(10.0, prec); double result = floor(num * tmp + 0.5); result /= tmp; return result; }是导致您数字古怪的原因。箩冀娥
更高精度的类型。 编辑：对不起，我没有注意到您得到的是完全不正确的结果。另一个答案是，这是由于转换为“ 7”溢出所致。铰齐插
#include <ieee754.h> #include <iostream> #include <iomanip> double round_double(double d, int precision) { ieee754_double* p = reinterpret_cast<ieee754_double*>(&d); std::cout << \"mantissa 0:\" << std::hex << p->ieee.mantissa0 << \", 1:\" << p->ieee.mantissa1 << \'\\n\'; unsigned mask0 = precision < 20 ? 0x000FFFFF << (20 - precision) : 0x000FFFFF; unsigned mask1 = precision < 20 ? 0 : precision == 53 ? 0xFFFFFFFF : 0xFFFFFFFE << (32 + 20 - precision); std::cout << \"masks 0:\" << mask0 << \", 1: \" << mask1 << \'\\n\'; p->ieee.mantissa0 &= mask0; p->ieee.mantissa1 &= mask1; std::cout << \"mantissa\' 0:\" << p->ieee.mantissa0 << \", 1:\" << p->ieee.mantissa1 << \'\\n\'; return d; } int main() { double num = 10.123456789987654321; for (int prec = 1; prec <= 53; ++prec) std::cout << std::setiosflags(std::ios::showpoint | std::ios::fixed) << std::setprecision(60) << \"round_double(\" << num << \", \" << prec << \") = \" << round_double(num, prec) << std::endl; }