c++ - Long double 没有提供所需的精度

Question

我正在读取一个文件，该文件的值以指数值的形式精确到 7 位数字

4.81591e-007

5.17968e-007

5.03954e-008

-8.30735e-007

虽然我得到的值只有 5 点精度

0.000000

0.000001

0.000000

-0.000001

-0.000002

代码如下，是

#include<iostream>
#include<vector>
#include<utility> 
#include<algorithm>
#include<map>
#include<cmath>
#include<sstream>
#include<string>
#include<queue>
#include<cstdlib>
#include<cstdio>
#include<fstream>
using namespace std;    
int main()
{
    FILE *fp2;
    fp2=fopen("output","w");
    int i=0;
    ifstream myFile;
    myFile.open("sruthi_DATA_.txt");
    vector<long double>numberlist;
    long double number;
    while(myFile >> number){    //
            numberlist.push_back(number);
            i++;
    }
    int j;
    for(j=0;j<i;j++)
    {
        cout.precision(10);
        cout<<numberlist[j]<<fixed<<endl;
        fprintf(fp2,"%Lf\n",numberlist[j]);
    }
    fclose(fp);
    fclose(fp2);

    return 0;

}

Answer 1

尝试将值打印为固定

http://www.cplusplus.com/reference/iomanip/setprecision/
http://www.cplusplus.com/reference/cstdio/printf/

 std::cout << std::fixed << std::setprecision(10) << value << "\n";

定义:

 fixed:       the value is represented with exactly as many digits in
              the decimal part as specified by the precision field
              (precision) and with no exponent part.

 scientific:  as many decimal digits as the precision field
              (precision). Finally, this notation always includes an
              exponential part consisting on the letter e followed by
              an optional sign and three exponential digits.

在 C 中打印:

  fprintf(fp2, "%.10f\n", value);

格式说明符的定义是:

  %[flags][width][.precision][length]specifier

  Precision: For a, A, e, E, f and F specifiers: this is the number
             of digits to be printed after the decimal point.

因此，要在 C++ 中实现与 C(%f) 相同的格式，您需要使用 std::fixed。