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;
}
}
答案 0 :(得分:1)
您需要在流中注入set precision:
file1&lt;&lt; setprecision(15)&lt;&lt;