我有一個程序,做一個辛ODE集成(物理/數學),我想將時間序列導出到一個.dat文件>然而,寫入的數字dat文件只有6位精度。我寫了setprecision(15);寫之前,但它沒有改變。我還張貼代碼的一部分,沒有實際的ODE求解:C++如何寫一個.dat文件的雙精度的全精度
#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;
}
所以,當我打開文件1(它沒有命名文件1)裏面的值是6位數字。我怎樣才能寫完整的16位準確數字?謝謝。 對於完整性我也發表名爲虛空功能:
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;
}
}
您的個人信息的定義將導致如果您曾經用過Solaris long double,那麼您會遇到麻煩:它不夠精確。 –
是的,這是真的,我不再使用它了!謝謝。 –