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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