使用固定点符号将double转换为字符串，没有尾随零和witout sprintf [英] Convert double to string with fixed point notation, no trailing zeroes and witout sprintf

问题描述

这个问题已被问过几次，但所有答案都涉及sprintf或涉及手动删除尾部零。真的没有更好的方法吗？是不可能实现这个与 std :: stringstream ？

This question has been asked a couple of times, but all the answers either refer to sprintf or involve deleting the trailing zeroes manually. Is there really no better way? is it not possible to achieve this with std::stringstream?

推荐答案

首先计算十进制之前和之后有多少位数：

First you calculate how many potential digits you have before and after the decimal:

int digits_before = 1 + (int)floor(log10(fabs(value)));
int digits_after = std::numeric_limits<double>::digits10 - digits_before;

然后你会发现这些数字中有多少是零：

Then you find out how many of those digits are zeros:

double whole = floor(pow(10, digits_after) * fabs(value) + 0.5);
while (digits_after > 0 && (whole/10.0 - floor(whole/10.0)) < 0.05)
{
    --digits_after;
    whole = floor(whole / 10.0 + 0.5);
}
if (digits_after < 0) digits_after = 0;

现在，您可以使用 std :: setprecision ：

Now you have a value you can use with std::setprecision:

std::stringstream ss;
ss << std::fixed << std::setprecision(digits_after) << value;

最终这是很多工作和重复的字符串转换的努力，一般只是转换为字符串并删除尾随零。没有，没有简单的格式化选项来做到这一点，你必须以硬的方式或根本不做。

Ultimately this is a lot of work and duplicates effort that the string conversion does anyway, which is why people generally just convert to a string and remove the trailing zeros. And no, there's no simple formatting option to do this, you have to do it the hard way or not at all.

查看上述代码在操作： http://ideone.com/HAk55Y

See the above code in action: http://ideone.com/HAk55Y

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