双转换为字符串有n位小数不尾随零 [英] Convert double to string with n decimal places without trailing zeros

问题描述

1）双D = 1.234567899; 结果
转换这一数字字符串8位小数不截断。
因此，预期成果是1.23456789，截短最后9。

1)double d = 1.234567899;
Convert this number to string with 8 decimal places without truncation. So,expected output is "1.23456789", truncating last 9.

和

2）如果d = 1.2345699; 结果
所以解决方案应该没有超出place.expected 8小数0追加输出1.2345699。

2)if d = 1.2345699;
so Solution should not append 0 upto 8th decimal place.expected output "1.2345699".

我已经尝试了多种解决方案，结束了stringstream的C ++类。第二个问题是解决了，但是第一个仍然存在。

I have tried many solutions,ended up with stringstream c++ class. 2nd problem is solved but first one still persist.

有什么办法来实现输出？

Is there any way to achieve the output?

先谢谢了。

推荐答案

如果要截断字符串重新presentation的一部分，无需四舍五入，你需要做的是手动：

If you want to truncate part of the string representation without rounding, you need to do that manually:

#include <iostream>
#include <sstream>
#include <string>
#include <iomanip>
#include <limits>

int main()
{
        std::stringstream s;
        double d = 1.234567899;

        // print it into sstream using maximum precision
        s << std::fixed << std::setprecision(std::numeric_limits<double>::digits10) << 1.234567899;
        std::string res = s.str();

        // Now the res contains something like 1.234567899000000
        // so truncate 9000000000 by hand

        size_t dotIndex = res.find(".");

        std::string final_res = res.substr(0, dotIndex + 9);

        std::cout << final_res << std::endl;

        return 0;
}

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