将double转换为带有n个小数位的字符串，不带尾随零 [英] Convert double to string with n decimal places without trailing zeros

问题描述

1） double d = 1.234567899;

将此数字转换为带有8位小数位数的字符串，不要截断。
所以，预期输出为1.23456789，截断最后9个。

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

if d = 1.2345699;

因此，解决方案不应将0添加到小数点后第8位。预期输出1.2345699。

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

我试过很多解决方案，最后用stringstream c ++类。第二个问题已解决，但第一个问题仍然存在。

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

有任何方法可以实现输出吗？

Is there any way to achieve the output?

推荐答案

如果要截断字符串表示形式的一部分而不舍入，则需要手动执行：

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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