C ++如何将完整的双精度写入.dat文件

时间:2015-07-12 11:39:36

标签: c++ export double

我有一个程序可以进行Symplectic ODE集成(物理/数学),我想将时间序列导出到.dat文件>但是,在dat文件中写入的数字只有6位精度。我写了setprecision(15);在写作之前,它没有任何改变。我还发布了一部分代码,没有实际的ODE求解器:

#include <iostream>
#include <ctime>         
#include <cstdlib>
#include <string>
#include <sstream>             
#include <iomanip>             
#include <cmath>              
#define pi 3.14159265358979
using namespace std;

int main(int argc, char **argv) {
// many stuff here, probably irrelevant,

ostringstream osE, osb, ospx;
    osb<<b; // using this, I can use some numbers into the file's name
    osE<<E;
    ospx<<px0;
filenamex = "Antidot_v4_x(t)_E=" + osE.str() + "_px0=" + ospx.str() + "_b=" + osb.str() + ".dat";
 ofstream file1( filenamex.c_str() );
file1<<t<<"\t"<<x0<<endl;
while(i<=N){
        i++;
        McLachanAtela(x, y, px, py, h);   
// Does the 4-step ODE solver. x are initial values and after the 
// function call x are final values after h time

        setprecision(15);
        file1<<t<<"\t"<<x<<endl; //I use this to write values in file
    }
 return 0;
}

因此,当我打开file1(它没有命名为file1)时,其中的值是6位数字。如何写出完整的16位数字准确数字?谢谢。 为了完整起见,我还发布了名为:

的void函数
void McLachanAtela (double& previousx, double& previousy, double& previouspx, double& previouspy, double timestep){
    // Atela Coefficients
    double c[4]={0.134496199277431089, -0.224819803079420805, 0.756320000515668291, 0.334003603286321425};
    double d[4]={0.515352837431122936, -0.085782019412973646, 0.411583023616466525, 0.128846158365384185};
    // Symplectic Algorithm (at dimensionless form)
    for(int j=0; j<4; j++){  
                                                        //this is the derivative of the potential :
        previouspx = previouspx - d[j]*timestep*b*pi*sin(pi*previousx)*cos(pi*previousy)*(pow(cos(pi*previousx)*cos(pi*previousy),b-1)); 
        previouspy = previouspy - d[j]*timestep*b*pi*sin(pi*previousy)*cos(pi*previousx)*(pow(cos(pi*previousx)*cos(pi*previousy),b-1));  
        previousx = previousx + c[j]*previouspx*timestep;
        previousy = previousy + c[j]*previouspy*timestep;
        //cout<<" dpx = "<<d[j]*timestep*b*pi*sin(pi*previousx)*cos(pi*previousy)*(pow(cos(pi*previousx)*cos(pi*previousy),b-1))<<endl;

    }
}

1 个答案:

答案 0 :(得分:1)

您需要在流中注入set precision:

file1&lt;&lt; setprecision(15)&lt;&lt;