Question

我需要一些使用C ++流的输出格式化帮助。我想打印固定小数点和最多2个尾随位置的数字。我尝试过以下方法：

#include <iostream>
#include <iomanip>
using namespace std;

int main(int argc, char **argv)
{
  float testme[] = { 0.12345, 1.2345, 12.345, 123.45, 1234.5, 12345 };

  std::cout << std::setprecision(2) << std::fixed;

  for(int i = 0; i < 6; ++i)
  {
      std::cout << testme[i] << std::endl;
  }

  return 0;
}

输出结果为：

但我想要

我可以在不使用其他字符串操作的情况下实现此目的吗？

Answer 1

我不知道适合的操纵者，因此你可以使用：

#include <iostream>
#include <iomanip>
#include <cmath>

template <typename T>
struct Fixed
{
    const T& value;
    const unsigned precision;
    const T significant;

    Fixed(const T& value, unsigned precision)
    : value(value), precision(precision), significant(std::pow(10, precision))
    {}

    void write(std::ostream& stream) const {
        // Adjust stream settings
        std::ostream::char_type restore_fill = stream.fill('0');
        std::ios_base::fmtflags restore_flags = stream.setf(
            std::ios_base::fixed, std::ios_base::floatfield);
        std::streamsize restore_precision = stream.precision(0);

        // Split the floating point into an integral and rounded fractional part
        T integral;
        unsigned long fractional = std::round(significant * std::modf(value, &integral));

        // Determine the length of the fractional part
        unsigned digits = precision;
        while(fractional && fractional % 10 == 0) {
            fractional /= 10;
            --digits;
        }

        // Carry over to the integral part
        if( ! digits && fractional) {
            integral += 1;
            fractional = 0;
        }

        // Output
        stream << integral;
        if(fractional) {
            stream <<  '.' << std::setw(digits) << fractional;
        }

        // Restore stream settings
        stream.precision(restore_precision);
        stream.flags(restore_flags);
        stream.fill(restore_fill);
    }
};

template <typename T>
inline Fixed<T> fixed(const T& value, unsigned precision) {
    return Fixed<T>(value, precision);
}

template <typename T>
inline std::ostream& operator << (std::ostream& stream, const Fixed<T>& value) {
    value.write(stream);
    return stream;
}

int main(int argc, char **argv)
{
  float testme[] = { 0.12345, 1.2345, 12.345, 123.45, 1234.5, 12345 };
  for(int i = 0; i < 6; ++i)
  {
      std::cout << fixed(testme[i], 2) << std::endl;
  }

  return 0;
}

Answer 2

这有效（http://ideone.com/CFcVhu），但不是那么漂亮......

#include <iostream>
#include <iomanip>
using namespace std;

int main(int argc, char **argv)
{
  float testme[] = { 0.12345, 1.2345, 12.345, 123.45, 1234.5, 12345 };

  //std::cout << std::setprecision(2) << std::fixed;

  for(int i = 0; i < 6; ++i)
  {
      std::cout << ((int)(testme[i]*100.0))/100.0f << std::endl;
  }

  return 0;
}

Answer 3

换句话说，如果它小于1000（总是正？），你想用2位小数表示一个数字，如果它小于10000，你想要1位小数，否则为0。嗯，怎么能把这个编成法呢？

限制固定输出的尾随零数

3 个答案: