如何在没有科学计数法或尾随零的情况下将浮点数输出到cout? [英] How to output float to cout without scientific notation or trailing zeros?
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问题描述
在C ++中没有科学计数法或尾随零的输出浮点数的最优雅方法是什么?
What is the most elegant way to output a floating point number in C++ with no scientific notation or trailing zeros?
float a = 0.000001f;
float b = 0.1f;
cout << "a: " << a << endl; // 1e-006 terrible, don't want sci notation.
cout << "b: " << b << endl; // 0.1 ok.
cout << fixed << setprecision(6);
cout << "a: " << a << endl; // 0.000001 ok.
cout << "b: " << b << endl; // 0.100000 terrible, don't want trailing zeros.
推荐答案
我不确定最优雅的方式
I am not sure about the "most elegant way" but here's one way.
#include <iostream>
#include <iomanip>
#include <sstream>
using namespace std ;
string fix( float x, int p )
{
ostringstream strout ;
strout << fixed << setprecision(p) << x ;
string str = strout.str() ;
size_t end = str.find_last_not_of( '0' ) + 1 ;
return str.erase( end ) ;
}
int main()
{
float a = 0.000001f ;
float b = 0.1f ;
cout << "a: " << fix( a, 6 ) << endl; // 0.000001 ok.
cout << "b: " << fix( b, 6 ) << endl; // 0.1 ok.
return 0;
}
您也许可以创建自己的I / O机械手,如果您需要大量此类输出的话。可以说这比较优雅,但实现方式可能相似。
You could perhaps create your own I/O manipulator if you need to to a lot of this kind of output. That is arguably more elegant, but the implementation could be similar.
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