C ++和Fortran的精度不同 [英] Different precision in C++ and Fortran

问题描述

对于我正在使用C ++编写的一个非常简单的函数的项目：

Fne（x）= 0.124 * x * x ，问题是当我计算<$ p
$ b

的函数值 x = 3.8938458092314270 与Fortran 77和C ++语言，我有不同的精度。

对于Fortran，我得到 Fne（x）= 1.8800923323458316 和C ++我得到 Fne（x）= 1.8800923630725743 。对于这两种语言，Fne函数被编码为双精度值，并且还返回双精度值。

C ++代码：

  double FNe（double X）{
 double FNe_out; 
 FNe_out = 0.124 * pow（X，2.0）; 
返回FNe_out; 
} 
  

Fortran代码：

  real * 8 function FNe（X）
 implicit real * 8（ah，oz）
 FNe = 0.124 * X * X 
 return 
 end 
  

你能帮我找一下这个区别吗？

解决方案

一个不同的来源是C ++和Fortran的默认处理常数，例如 0.124 。默认情况下，Fortran会将其视为单精度浮点数（几乎可以使用任何计算机和编译器组合），而C ++将其视为双精度fp数。

在Fortran中，您可以指定一个fp数字（或任何其他内在数字常量）的类，并且没有任何编译器选项可以更改最可能的默认行为）通过后缀 kind-selector 这样

  0.124_8 
  

尝试一下，看看有什么结果。

哦而在写作的时候，为什么你写的是Fortran，就像1977？对于所有其他Fortran专家来说，是的，我知道， * 8 和 _8 不是最佳做法，我现在还没有时间来扩展这个。

For a project I'm working on I've coded in C++ a very simple function :

Fne(x) = 0.124*x*x, the problem is when i compute the value of the function

for x = 3.8938458092314270 with both Fortran 77 and C++ languages , i got different precison.

For Fortran I got Fne(x) = 1.8800923323458316 and for C++i got Fne(x) = 1.8800923630725743. For both languages, the Fne function is coded for double precision values, and return also double precision values.

C++ code:

double FNe(double X) {
    double FNe_out;
    FNe_out = 0.124*pow(X,2.0);
    return FNe_out;
}

Fortran code:

  real*8 function FNe(X)
  implicit real*8 (a-h,o-z)
  FNe = 0.124*X*X
  return
  end

Can you please help me to find where this difference is from?

解决方案

One source of difference is the default treatment, by C++ and by Fortran, of literal constants such as your 0.124. By default Fortran will regard this as a single-precision floating-point number (on almost any computer and compiler combination that you are likely to use), while C++ will regard it as a double-precision f-p number.

In Fortran you can specify the kind of a f-p number (or any other intrinsic numeric constant for that matter and absent any compiler options to change the most-likely default behaviour) by suffixing the kind-selector like this

0.124_8

Try that, see what results.

Oh, and while I'm writing, why are you writing Fortran like it was 1977 ? And to all the other Fortran experts hereabouts, yes, I know that *8 and _8 are not best practice, but I haven't the time at the moment to expand on all that.

这篇关于C ++和Fortran的精度不同的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！